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Calculus AB Practice Exam and Notes

Effective Fall 2016

Important Note

This Practice Exam is provided by the College Board for AP Exam preparation. Teachers are permitted to download the materials and make copies to use with their students in a classroom setting only. To maintain the security of this exam, teachers should collect all materials after their administration and keep them in a secure location.

Exams may not be posted on school or personal websites, nor electronically redistributed for any reason. Further distribution of these materials outside of the secure College Board site disadvantages teachers who rely on uncirculated questions for classroom testing. Any additional distribution is in violation of the College Board’s copyright policies and may result in the termination of Practice Exam access for your school as well as the removal of access to other online services such as the AP Teacher Community and Online Score Reports.

UpdatedRationale C for question 30 has been corrected.

Calculus AB Practice Exam 2

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http://www.collegeboard.org


Contents

Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

I. Practice ExamExam Content and Format . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

Administering the Practice Exam . . . . . . . . . . . . . . . . . . . . . . . . . 7

Answer Sheet for Multiple-Choice Section . . . . . . . . . . . . . . . . . . 10

AP® Calculus AB Practice Exam . . . . . . . . . . . . . . . . . . . . . . . . . . 11

II. Notes on the Practice ExamIntroduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57

Multiple-Choice Section . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59

Answers to Multiple-Choice Questions . . . . . . . . . . . . . . . . . . . . . 108

Free-Response Section . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 109

Contact Us . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121


IntroductionBeginning in May 2017, the AP Calculus AB Exam will measure students’ ability to apply strategies and techniques to accurately solve diverse types of problems. The revised exam will feature the same number of questions and total allotted time, but the distribution of questions and relative timing have been adjusted based on feedback from teachers and administrators, and multiple-choice questions now have four answer choices instead of five.

Part I of this publication is the AP Calculus AB Practice Exam. This will mirror the look and feel of an actual AP Exam, including instructions and sample questions. However, these exam questions have never been administered as an operational exam, and, therefore, statistical analysis is not available. The purpose of this section is to provide educators with sample exam questions that accurately reflect the composition/design of the revised exam and to offer these questions in a way that gives teachers the opportunity to test their students in an exam situation that closely resembles the actual exam administration.

Important: Final instructions for every AP Exam are published in the AP Exam Instructions book. Please reference that publication, which is posted at www.collegeboard.org/apexaminstructions in March and included in schools’ exam shipments, for the final instructions and format of this AP Exam.

Part II is the Notes on the AP Calculus AB Practice Exam. This section offers detailed explanations of how each question in the practice exam links back to the curriculum framework in order to provide a clear link between curriculum and assessment. The multiple-choice rationales explain the correct answer and incorrect options. Scoring information is provided for the free-response section.

How AP Courses and Exams Are DevelopedAP courses and exams are designed by committees of college faculty and AP teachers who ensure that each AP course and exam reflects and assesses college-level expectations. These committees define the scope and expectations of the course, articulating through a curriculum framework what students should know and be able to do upon completion of the AP course. Their work is informed by data collected from a range of colleges and universities to ensure that AP course work reflects current scholarship and advances in the discipline.

These same committees are also responsible for designing and approving exam specifications and exam questions that clearly connect to the curriculum framework. The AP Exam development process is a multiyear endeavor; all AP Exams undergo extensive review, revision, piloting, and analysis to ensure that questions are high quality and fair and that the questions comprise an appropriate range of difficulty.

Throughout AP course and exam development, the College Board gathers feedback from secondary and postsecondary educators. This feedback is carefully considered to ensure that AP courses and exams provide students with a college-level learning experience and the opportunity to demonstrate their qualifications for advanced placement and college credit upon college entrance.

http://www.collegeboard.org/apexaminstructions


Methodology Guiding the RevisionThe course and the exam are conceived and developed using similar methodologies. The course is designed using the principles of Understanding by Design, and the exam is designed and developed using the similarly principled evidence-centered design approach. Both processes begin by identifying the end goals that identify what students should know and be able to do by the end of their AP experience. These statements about students’ knowledge and abilities, along with descriptions of the observable evidence that delineate levels of student performance, serve simultaneously as the learning objectives for the course and the targets of measurement for the exam. The course and exam, by design, share the same foundation.

Course Development

Each committee first articulates its discipline’s high-level goals before identifying the course’s specific learning objectives. This approach is consistent with “backward design” — the practice of developing curricula, instruction, and assessments with the end goal in mind. The learning objectives describe what students should know and be able to do, thereby providing clear instructional goals as well as targets of measurement for the exam.

Exam Development

Exam development begins with the committee making decisions about the overall nature of the exam. How will the learning objectives for the course be assessed? How will the course content and skills be distributed across the exam? How many multiple-choice questions should there be? How many free-response questions should be included? How much time will be devoted to each section? Answers to these questions become part of the exam specifications.

With the exam specifications set, assessment specialists design questions that conform to these specifications. The committee reviews every exam question for alignment with the curriculum framework, accuracy, and a number of other criteria that ensure the integrity of the exam.

Exam questions are then piloted in AP classrooms to determine their statistical properties. Questions that have been approved by the committee and piloted successfully are included in an exam. When an exam is assembled, the committee conducts a final review to ensure overall conformity with the specifications.

How AP Exams Are ScoredThe exam scoring process, like the course and exam development process, relies on the expertise of both AP teachers and college faculty. While multiple-choice questions are scored by machine, the free-response questions and, as applicable, through-course performance assessments, are scored by college faculty and expert AP teachers at the annual AP Reading. Most of the Reading occurs in face-to-face settings, while a small portion are scored online while the face-to-face Reading is taking place.


AP Exam Readers are thoroughly trained, and their work is monitored throughout the Reading for fairness and consistency. In each subject, a highly respected college faculty member fills the role of Chief Reader, who, with the help of AP Readers in leadership positions, maintains the accuracy of the scoring standards. Scores on the free-response questions and performance assessments are weighted and combined with the weighted results of the computer-scored multiple-choice questions to yield the weighted composite score, and this composite score is converted into an AP exam score of 5, 4, 3, 2, or 1.

The score-setting process is both precise and labor intensive, involving numerous psychometric analyses of the results of a specific AP Exam in a specific year and of the particular group of students who took that exam. Additionally, to ensure alignment with college-level standards, part of the score-setting process involves comparing the performance of AP students with the performance of students enrolled in comparable courses in colleges throughout the United States. In general, the AP composite score points are set so that the lowest raw score needed to earn an AP score of 5 is equivalent to the average score among college students earning grades of A in the college course. Similarly, AP Exam scores of 4 are equivalent to college grades of A–, B+, and B. AP Exam scores of 3 are equivalent to college grades of B–, C+, and C.

Using and Interpreting AP ScoresThe extensive work done by college faculty and AP teachers in the development of the course and the exam and throughout the scoring process ensures that AP Exam scores accurately represent students’ achievement in the equivalent college course. While colleges and universities are responsible for setting their own credit and placement policies, AP scores signify how qualified students are to receive college credit and placement:

AP Score Recommendation

5 Extremely well qualified

4 Well qualified

3 Qualified

2 Possibly qualified

1 No recommendation

Additional ResourcesVisit apcentral.collegeboard.org for more information about the AP Program.

http://apcentral.collegeboard.org


AP Calculus AB Practice Exam

Exam Content and Format

The 2017 AP Calculus AB Exam is 3 hours and 15 minutes in length. There are two sections:

• Section I is 1 hour, 45 minutes and consists of 45 multiple-choice questions in two separately-timed parts, accounting for 50 percent of the final score. Part A consists of 30 questions in 60 minutes and does not allow the use of a calculator. Part B consists of 15 questions in 45 minutes and requires the use of a graphing calculator.

• Section II is 1 hour, 30 minutes and consists of 6 free-response questions in two separately-timed parts, accounting for 50 percent of the final score. Part A consists of 2 questions in 30 minutes and requires the use of a graphing calculator. Part B consists of 4 questions in 60 minutes and does not allow the use of a calculator. During the timed portion for Part B, students are permitted to continue to work on questions in Part A, but they are not allowed to use a calculator during this time.

Administering the Practice Exam

This section contains instructions for administering the AP Calculus AB Practice Exam. You may wish to use these instructions to create an exam situation that resembles an actual administration. If so, read the indented, boldface directions to the students; all other instructions are for administering the exam and need not be read aloud. Before beginning testing, have all exam materials ready for distribution. These include exam booklets and answer sheets.

Graphing calculators are required to answer some of the questions on the AP Calculus Exams. Before starting the exam administration, make sure each student has a graphing calculator from the approved list at https://apstudent.collegeboard.org/apcourse/ap-calculus-ab/calculator-policy or on AP Central. During the administration of Section I, Part B, and Section II, Part A, students may have no more than two graphing calculators on their desks. Calculators may not be shared. Calculator memories do not need to be cleared before or after the exam. Since graphing calculators can be used to store data, including text, teachers should monitor that students are using their calculators appropriately.

Section I of the Practice Exam should be completed using a No. 2 pencil to simulate an actual administration. Students may use a No. 2 pencil or a pen with black or dark blue ink to complete Section II.

Instructions for the Section II free-response questions are included in this publication. It is important to share these with students and ask them to read these instructions carefully at the beginning of the administration of Section II. Timing for Section II should begin after you have given sufficient time to read these instructions.

https://apstudent.collegeboard.org/apcourse/ap-calculus-ab/calculator-policy


SECTION I, Multiple-Choice Questions — Part A (no calculator allowed) and Part B (graphing calculator required)

When you are ready to begin Section I, say:

Section I is the multiple-choice portion of the exam. Mark all of your responses on your answer sheet, one response per question. If you need to erase, do so carefully and completely. Your score on the multiple-choice section will be based solely on the number of questions answered correctly.

Section I is divided into two parts. Each part is timed separately, and you may work on each part only during the time allotted for it. Calculators are not allowed in Part A. Please put all of your calculators under your chair. Are there any questions?…

You have 60 minutes for Part A. Part A questions are numbered 1 through 30. Open your Section I booklet to Part A and begin.

Note Start Time here______ . Note Stop Time here______ . Check that students are marking their answers in pencil on the answer sheets and that they are not looking beyond Part A. The line of A’s at the top of each page will assist you in monitoring students’ work. After 50 minutes, say:

There are 10 minutes remaining in Part A.

After 10 minutes, say:

Stop working on Part A. Graphing calculators are required for Part B. You may get your calculators from under your chair and place them on your desk. You have 45 minutes for Part B. Part B questions are numbered 76 through 90. Open your Section I booklet to Part B and begin.

Note Start Time here______ . Note Stop Time here______ . Check that students are marking their answers in pencil on the answer sheets and are now working on Part B. The large B’s in an alternating shaded pattern at the top of each page will assist you in monitoring their work. After 35 minutes, say:

There are 10 minutes remaining in Part B.


Stop working and close your exam booklet. I will now collect your Section I booklet and your answer sheet. Put your exam booklet and your answer sheet on your desk, face up. Remain in your seat, without talking, while the exam materials are collected.

Collect a Section I booklet and answer sheet from each student.

There is a 10-minute break between Sections I and II.


SECTION II: Free-Response Questions — Part A (graphing calculator required) and Part B (no calculator allowed)

After the break, say:

Section II is the free-response portion of the exam. It also has two parts that are timed separately. You are responsible for pacing yourself, and may proceed freely from one question to another within each part. Graphing calculators are required for Part A, so you may keep your calculators on your desk. You must write your answers in the appropriate space in the exam booklet using a No. 2 pencil or a pen with black or dark blue ink. Do not begin Part B until you are told to do so.

Please read the instructions for the free-response section, paying careful attention to the bulleted statements. Look up when you have finished….

Are there any questions?...

You have 30 minutes to answer the questions in Part A. Open your Section II booklet and begin.

Note Start Time here______ . Note Stop Time here______ . Check that students are working on Part A only and writing their answers in their exam booklets using pencils or pens with black or dark blue ink. The pages for Part A questions are marked with large 1s or 2s at the top of each page to assist you in monitoring their work. After 20 minutes, say:

There are 10 minutes remaining in Part A.


Stop working on Part A. Calculators are not allowed for Part B. Please put all of your calculators under your chair….

You have 60 minutes for Part B. During this time you may go back to Part A, but you may not use your calculator. Remember to show your work, and write your answer to each part of each question in the appropriate space in the exam booklet. Are there any questions?...

Open your Section II booklet to Part B and begin.

Note Start Time here______ . Note Stop Time here______ . After 50 minutes, say:

There are 10 minutes remaining in Part B.


Stop working and close your exam booklet. Put your exam booklet on your desk, face up. Remain in your seat, without talking, while the exam materials are collected.

Collect a Section II booklet from each student. Then say:

The exam is now completed.


Name:

AP® Calculus AB Answer Sheet

for Multiple-Choice Section

No. Answer

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

17

18

19

20

21

22

23

24

25

26

27

28

29

30

No. Answer

76

77

78

79

80

81

82

83

84

85

86

87

88

89

90


At a GlanceTotal Time1 hour, 45 minutes

Number of Questions45

Percent of Total Score50%

Writing InstrumentPencil required

Part ANumber of Questions30

Time60 minutes

Electronic DeviceNone allowed

Part BNumber of Questions15

Time45 minutes

Electronic DeviceGraphing calculatorrequired

InstructionsSection I of this exam contains 45 multiple-choice questions. For Part A, fill in only theboxes for numbers 1 through 30 on the answer sheet. For Part B, fill in only the boxes fornumbers 76 through 90 on the answer sheet.

Indicate all of your answers to the multiple-choice questions on the answer sheet. Nocredit will be given for anything written in this exam booklet, but you may use the bookletfor notes or scratch work. After you have decided which of the suggested answers is best,place the letter of your choice in the corresponding box on the answer sheet. Give onlyone answer to each question.

Use your time effectively, working as quickly as you can without losing accuracy. Do notspend too much time on any one question. Go on to other questions and come back tothe ones you have not answered if you have time. It is not expected that everyone willknow the answers to all of the multiple-choice questions.

Your total score on the multiple-choice section is based only on the number of questionsanswered correctly. Points are not deducted for incorrect answers or unansweredquestions.

DO NOT OPEN THIS BOOKLET UNTIL YOU ARE TOLD TO DO SO.

AP® Calculus AB ExamSECTION I: Multiple Choice


GO ON TO THE NEXT PAGE.

CALCULUS AB

SECTION I, Part A

Time—60 minutes

Number of questions—30

NO CALCULATOR IS ALLOWED FOR THIS PART OF THE EXAM.

Directions: Solve each of the following problems, using the available space for scratch work. After examining theform of the choices, decide which is the best of the choices given and place the letter of your choice in thecorresponding box on the answer sheet. No credit will be given for anything written in this exam booklet. Do notspend too much time on any one problem.

In this exam:

(1) Unless otherwise specified, the domain of a function f is assumed to be the set of all real numbers x for which

f x( ) is a real number.

(2) The inverse of a trigonometric function f may be indicated using the inverse function notation f 1− or with the

prefix “arc” (e.g., x xsin arcsin1 =− ).

A A A A A A A A A A A A A A A A A A A A A A A A A A A A A A A




1. ∫ x x dx21

3 2 =( − )−

(A)2

3− (B)

4

3(C) 8 (D) 12

2. If f x e x 1x2 3= +( ) ( ), then f 2¢ =( )

(A) e6 4 (B) e21 4 (C) e24 4 (D) e30 4


3. Let f be a differentiable function such that f 2 4=( ) and f 21

2¢ =( ) − . What is the approximation for f 2.1( )

found by using the line tangent to the graph of f at x 2= ?

(A) 2.95 (B) 3.95 (C) 4.05 (D) 4.1

4. Let g be the function defined by g x x x44 3= +( ) . How many relative extrema does g have?

(A) Zero (B) One (C) Two (D) Three




5. The velocity of a particle moving along the x-axis is given by v t t2 2=( ) − for time t 0> . What is the average

velocity of the particle from time t 1= to time t 3= ?

(A) 4− (B) 3− (C)7

3− (D)

7

3

6. On a certain day, the rate at which material is deposited at a recycling center is modeled by the function R,

where R t( ) is measured in tons per hour and t is the number of hours since the center opened. Using a

trapezoidal sum with the three subintervals indicated by the data in the table, what is the approximate number of

tons of material deposited in the first 9 hours since the center opened?

(A) 68 (B) 70.5 (C) 85 (D) 136




7. What is the total area of the regions between the curves y x x6 182= − and y x6= − from x 1= to x 3= ?

(A) 4 (B) 12 (C) 16 (D) 20

8. The function g is defined by g x x bx2= +( ) , where b is a constant. If the line tangent to the graph of g

at x 1= − is parallel to the line that contains the points 0, 2( − ) and 3, 4( ), what is the value of b ?

(A) 1− (B) 2 (C)5

2(D) 4




f xx

xx

x

for 0

0 for 0

= π

=( )

⎧⎨⎪⎪⎪

⎩⎪⎪⎪

9. The function f is defined above. The value of ∫ f x dx5

3( )

−is

(A) 2− (B) 2 (C) 8 (D) nonexistent

10. Let g be a continuous function. Using the substitution u x2 1= − , the integral ∫ g x dx2 12

3( − ) is equal to

which of the following?

(A) ∫ g u du2

3( ) (B) ∫ g u du

1

2 2

3( ) (C) ∫ g u du

3

5( ) (D) ∫ g u du

1

2 3

5( )






11. The graph of y f x= ( ) consists of a semicircle with endpoints at 2, 6( − ) and 12, 6( − ), as shown in the figure

above. What is the value of ∫ f x dx2

12( ) ?

(A)25

2

p− (B)

25

2

p(C) 60

25

2

p+− (D) 60

25

2

p−




12. An object moves along a straight line so that at any time t its acceleration is given by a t t6=( ) . At time t 0= ,

the object’s velocity is 10 and the object’s position is 7. What is the object’s position at time t 2= ?

(A) 22 (B) 27 (C) 28 (D) 35

13. If y x xcos ln 2= − ( ), thend y

dx

3

3 =

(A) xx

sin23−

(B) xx

sin23− −

(C) xx

sin13−

(D) xx

sin13− −




14. The graph of the function f, shown above, consists of three line segments. If the function g is an antiderivative

of f such that g 2 5=( ) , for how many values of c, where c0 6£ £ , does g c 3=( ) ?

(A) Zero (B) One (C) Two (D) Three

f xcx x

x x6 for 19 2 ln for 1

= ++

( )⎧⎨⎪⎪⎩⎪⎪

<≥

15. Let f be the function defined above, where c is a constant. If f is continuous at x 1= , what is the value of c ?

(A) 2 (B) 3 (C) 5 (D) 9




16. Which of the following could be a slope field for the differential equationdy

dxx y2= + ?

(A) (B)

(C) (D)




17. The function y e x5 7x3= +− is a solution to which of the following differential equations?

(A) y y3 15 0≤ ¢ =− −

(B) y y3 15 0≤ ¢ + =−

(C) y y 5 0≤ ¢ =− −

(D) y y 5 0≤ ¢ + =−

18. If f x xsin 1=( ) − , then f3

2¢⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟=

(A)6

p(B)

3

p(C)

4

7(D) 2




19.x x e e

x elim

3 3

x e

20 20( − ) − ( − )−→

is

(A) 0 (B) e20 319 − (C) e 320 − e (D) nonexistent

20. Let y f x= ( ) be a twice-differentiable function such that f 1 2=( ) anddy

dxy 33= + . What is the value of

d y

dx

2

2

at x 1= ?

(A) 12 (B) 66 (C) 132 (D) 165


21. The table above gives values of f, f ¢, g, and g ¢ for selected values of x. If h x f g x=( ) ( ( )), what is the value

of h 1¢( ) ?

(A) 19− (B) 14− (C) 7 (D) 9

22. Let y f x= ( ) be the particular solution to the differential equationdy

dx

x

y

1=

+with the initial condition

f 0 2=( ) − . Which of the following is an expression for f x( ) ?

(A) x x2 22 +− −

(B) x x2 22+ +−

(C) x x2 42 + +−

(D) x x2 42 + +




23. Let R be the shaded region bounded by the graph of y x= , the graph of y x 2= − , and the x-axis, as shownin the figure above. Which of the following gives the volume of the solid generated when R is revolved aboutthe x-axis?

(A) ∫ x x dx20

4 2p ( − ( − ) )

(B) ∫ x x dx20

4 2p ( − ( − ))

(C) ∫ ∫x dx x x dx20

2

2

4 2p p+ ( − ( − ) )

(D) ∫ ∫x dx x x dx20

2

2

4 2p p+ ( − ( − ))






24.x

e xlim

tan 3

3x x3 3

( − )

−−→is

(A) 0 (B)1

3(C)

1

2(D) nonexistent

25. Let f be a function with first derivative defined by f xx

x

3 62

2¢ =( )−

for x 0> . It is known that f 1 9=( ) and

f 3 11=( ) . What value of x in the open interval 1, 3( ) satisfies the conclusion of the Mean Value Theorem for f

on the closed interval 1, 3[ ] ?

(A) 6 (B) 3 (C) 2 (D) 1




26.x x

xdx

5

21

2 2

+=

⌠

⌡⎮⎮

− −

(A)3

2ln

4

3+− (B)

25

21− (C)

5

23 ln

3

4+ (D)

23

45

27. A hemispherical water tank, shown above, has a radius of 6 meters and is losing water. The area of the surface

of the water is A h h12 2p p= − square meters, where h is the depth, in meters, of the water in the tank. When

h 3= meters, the depth of the water is decreasing at a rate of1

2meter per minute. At that instant, what is the

rate at which the area of the water’s surface is decreasing with respect to time?

(A) 3p square meters per minute

(B) 6p square meters per minute

(C) 9p square meters per minute

(D) 27p square meters per minute



28. Consider a triangle in the xy-plane. Two vertices of the triangle are on the x-axis at 1, 0( ) and 5, 0( ), and a third

vertex is on the graph of y x xln 21

25= +( ) − for x

1

28£ £ . What is the maximum area of such a triangle?

(A)19

2

(B) 2 ln 2 9+

(C) 2 ln 4 8+

(D) 2 ln 16 2+

29. The function f is defined by f x x x4 23= + +( ) . If g is the inverse function of f and g 2 0=( ) , what is the

value of g 2¢( ) ?

(A)1

16− (B)

4

81− (C)

1

4(D) 4



30. Which of the following limits is equal to ∫ x dx2

5 2 ?

(A)∞

k

n nlim 2

1

n k

n

1

2

+=

⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟∑

→

(B)∞

k

n nlim 2

3

n k

n

1

2

+=

⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟∑

→

(C)∞

k

n nlim 2

3 1

n k

n

1

2

+=

⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟∑

→

(D)∞

k

n nlim 2

3 3

n k

n

1

2

+=

⎛⎝⎜⎜⎜

⎞⎠⎟⎟⎟∑

→

END OF PART A

IF YOU FINISH BEFORE TIME IS CALLED,YOU MAY CHECK YOUR WORK ON PART A ONLY.

DO NOT GO ON TO PART B UNTIL YOU ARE TOLD TO DO SO.



B B B B B B B B B


CALCULUS AB

SECTION I, Part B

Time—45 minutes

Number of questions—15

A GRAPHING CALCULATOR IS REQUIRED FOR SOME QUESTIONS ON THIS PART OF THE EXAM.

Directions: Solve each of the following problems, using the available space for scratch work. After examining theform of the choices, decide which is the best of the choices given and place the letter of your choice in thecorresponding box on the answer sheet. No credit will be given for anything written in this exam booklet. Do notspend too much time on any one problem.

In this exam:

(1) The exact numerical value of the correct answer does not always appear among the choices given. When thishappens, select from among the choices the number that best approximates the exact numerical value.

(2) Unless otherwise specified, the domain of a function f is assumed to be the set of all real numbers x for which

f x( ) is a real number.

(3) The inverse of a trigonometric function f may be indicated using the inverse function notation f 1− or with the

prefix “arc” (e.g., x xsin arcsin1 =− ).


B B B B B B B B B


76. To help restore a beach, sand is being added to the beach at a rate of s t t65 24 sin 0.3= +( ) ( ) tons per hour,

where t is measured in hours since 5:00 A.M. How many tons of sand are added to the beach over the 3-hour

period from 7:00 A.M. to 10:00 A.M.?

(A) 255.368 (B) 225.271 (C) 85.123 (D) 10.388

77. The graph of the function f is shown above. For what values of a does f xlim 0x a

=( )→

?

(A) 2 only

(B) 2 and 4

(C 0 and 2 only)

(D) 0, 1, and 2


B B B B B B B B B


78. The second derivative of a function f is given by f x x xsin 3 cos 2≤ =( ) ( ) − ( ). How many points of inflection

does the graph of f have on the interval x0 3< < ?

(A) One (B) Three (C Four) (D) Five

79. Over the time interval t0 5£ £ , a particle moves along the x-axis. The graph of the particle’s velocity, v,

is shown above. Over the time interval t0 5£ £ , the particle’s displacement is 3 and the particle travels a

total distance of 13. What is the value of ∫ v t dt2

4( ) ?

(A) 10− (B) 5− (C) 5 (D) 10


B B B B B B B B B


80. The temperature in a room at midnight is 20 degrees Celsius. Over the next 24 hours, the temperature changes

at a rate modeled by the differentiable function H, where H t( ) is measured in degrees Celsius per hour and tim

t is measured in hours since midnight. Which of the following is the best interpretation of ∫ H t dt0

6( ) ?

e

(A) The temperature of the room, in degrees Celsius, at 6:00 A.M.

(B) The average temperature of the room, in degrees Celsius, between midnight and 6:00 A.M.

(C) The change in the temperature of the room, in degrees Celsius, between midnight and 6:00 A.M.

(D) The rate at which the temperature in the room is changing, in degrees Celsius per hour, at 6:00 A.M.


B B B B B B B B B

(A) (B)

(C) (D)


81. The graph of f ¢, the derivative of the function f, is shown above. Which of the following could be the graphof f ?


B B B B B B B B B

82. Let f be the function with derivative given by f x xsin 32¢ =( ) ( − ). At what values of x in the interval

x3 3< <− does f have a relative maximum?

(A) 1.732− and 2.478 only

(B) 2.478− and 1.732 only

(C) 2.138− , 0, and 2.138

(D) 2.478− , 1.732− , 1.732, and 2.478

83. The graph of the function f is shown above. At what value of x does f have a jump discontinuity?

(A) 1 (B) 3 (C) 7 (D) 10



B B B B B B B B B

84. Let f be a differentiable function such that f 1 p=( ) and f x x 63¢ = +( ) . What is the value of f 5( ) ?

(A) 11.941 (B) 14.587 (C) 24.672 (D) 27.814

85. People are entering a building at a rate modeled by f t( ) people per hour and exiting the building at a rate

modeled by g t( ) people per hour, where t is measured in hours. The functions f and g are nonnegative and

differentiable for all times t. Which of the following inequalities indicates that the rate of change of the number

of people in the building is increasing at time t ?

(A) f t 0>( )

(B) f t 0¢ >( )

(C) f t g t 0>( ) − ( )

(D) f t g t 0¢ ¢ >( ) − ( )



B B B B B B B B B

86. The velocity of a particle moving along the x-axis is given by v t t ecos t=( ) − ( ) for t 0≥ . Which of the

following statements describes the motion of the particle at t 1= ?

(A) The particle is moving to the left with positive acceleration.

(B) The particle is moving to the right with positive acceleration.

(C) The particle is moving to the left with negative acceleration.

(D) The particle is moving to the right with negative acceleration.

87. A tire that is leaking air has an initial air pressure of 30 pounds per square inch (psi). The function t f p= ( )models the amount of time t, in hours, it takes for the air pressure of the tire to reach p psi. What are the units

for f p¢( ) ?

(A) hours (B) psi (C) psi per hour (D) hours per psi



B B B B B B B B B

88. The first derivative of the function f is defined by f xx e

x

2

0.7

x

2¢ =++

( )−

. On what intervals is f increasing?

(A) x1.384 0.264< <− − only

(B) x 0.633< − and x 0.319> only

(C) ∞ ∞x< <−

(D) There are no intervals on which f is increasing.

89. The table above shows selected values of a continuous function f. For x0 13£ £ , what is the fewest possible

number of times f x 4=( ) ?

(A) One (B) Two (C) Three (D) Four



B B B B B B B B B

90. The function h is defined on the closed interval 1, 3[ ]− . The graph of h¢, the derivative of h, is shown above.

The graph consists of two semicircles with a common endpoint at x 1= . Which of the following statements

about h must be true?

I. h h1 3=(− ) ( )

II. h is continuous at x 1= .

III. The graph of h has a vertical asymptote at x 1= .

(A) None (B) II only (C) I and II only (D) I and III only

END OF SECTION I

IF YOU FINISH BEFORE TIME IS CALLED,YOU MAY CHECK YOUR WORK ON PART B ONLY.

DO NOT GO ON TO SECTION II UNTIL YOU ARE TOLD TO DO SO.


DO NOT OPEN THIS BOOKLET OR BEGIN PART B UNTIL YOU ARE TOLD TO DO SO.

At a GlanceTotal Time1 hour, 30 minutes

Number of Questions6

Percent of Total Score50%

Writing InstrumentEither pencil or pen withblack or dark blue ink

WeightThe questions areweighted equally, butthe parts of a questionare not necessarilygiven equal weight.

Part ANumber of Questions2

Time30 minutes

Electronic DeviceGraphing calculatorrequired

Percent of Section II Score33.3%

Part BNumber of Questions4

Time60 minutes

Electronic DeviceNone allowed

Percent of Section II Score66.6%

InstructionsThe questions for Section II are printed in this booklet. Do not begin Part B until you aretold to do so. Write your solution to each part of each question in the space provided.Write clearly and legibly. Cross out any errors you make; erased or crossed-out work willnot be scored.

Manage your time carefully. During the timed portion for Part A, work only on thequestions in Part A. You are permitted to use your calculator to solve an equation, findthe derivative of a function at a point, or calculate the value of a definite integral.However, you must clearly indicate the setup of your question, namely the equation,function, or integral you are using. If you use other built-in features or programs, youmust show the mathematical steps necessary to produce your results. During the timedportion for Part B, you may continue to work on the questions in Part A without the useof a calculator.

For each part of Section II, you may wish to look over the questions before starting towork on them. It is not expected that everyone will be able to complete all parts of allquestions.• Show all of your work, even though a question may not explicitly remind you to do so.

Clearly label any functions, graphs, tables, or other objects that you use. Justificationsrequire that you give mathematical reasons, and that you verify the needed conditionsunder which relevant theorems, properties, definitions, or tests are applied. Your workwill be scored on the correctness and completeness of your methods as well as youranswers. Answers without supporting work will usually not receive credit.

• Your work must be expressed in standard mathematical notation rather than calculator

syntax. For example, �1

5x2 dx may not be written as fnInt(X2, X, 1, 5).

• Unless otherwise specified, answers (numeric or algebraic) need not be simplified. If youuse decimal approximations in calculations, your work will be scored on accuracy.Unless otherwise specified, your final answers should be accurate to three places afterthe decimal point.

• Unless otherwise specified, the domain of a function f is assumed to be the set of all

AP® Calculus AB ExamSECTION II: Free Response

real numbers x for which f(x) is a real number.



CALCULUS AB SECTION II, Part A

Time—30 minutes Number of problems—2

A GRAPHING CALCULATOR IS REQUIRED FOR THESE PROBLEMS.


111 11 11 1 1

1

Do not w

rite beyond this border. Do

not w

rite

beyo

nd th

is b

orde

r.

1. The depth of water in tank A, in inches, is modeled by a differentiable and increasing function h for£ £t0 10, where t is measured in minutes. Values of h t( ) for selected values of t are given in the table above.

(a) Use the data in the table to find an approximation for h 6¢( ). Show the computations that lead to your

answer. Indicate units of measure.

(b) Approximate the value of ∫ h t dt0

10( ) using a right Riemann sum with the four subintervals indicated by

the data in the table. Is this approximation greater than or less than ( )∫ h t dt0

10? Give a reason for your

answer.


111 11 11 1 1

1


Do

not w

rite

beyo

nd th

is b

orde

r.

Do not w

rite beyond this border.

(d) According to the model given in part (c), is the depth of the water in tank B increasing or decreasing attime t 6= ? Give a reason for your answer.

(c) The depth of water in tank B, in inches, is modeled by the function g t t3.2 17.5 sin 0.16( ) = + ( ) for

£ £t0 10, where t is measured in minutes. Find the average depth of the water in tank B over the

interval t0 10£ £ . Is this value greater than or less than the average depth of the water in tank A over the

interval £ £t0 10 ? Give a reason for your answer.


222 22 22 2 2

2

Do not w


not w

rite

beyo

nd th

is b

orde

r.

2. Particle Q moves along the x-axis so that its velocity at any time t is given by ( ) = −⎛

⎝⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟v t

t1 3 cos

5Q

2

, and its

acceleration at any time t is given by a tt t6

5sin

5Q

2

( ) =⎛

⎝⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟. The particle is at position x 2= at time t 0= .

(a) In the interval t0 5< < , when is the velocity of particle Q increasing? Give a reason for your answer.

(b) Find the position of particle Q at time t 3= .


222 22 22 2 2

2

Do not w


Do

not w

rite

beyo

nd th

is b

orde

r.

(c) A second particle, R, moves along the x-axis so that its position at any time t is given by a differentiablefunction ( )x tR , where x 1 4R( ) = and x 3 8R( ) = . Explain why there must be a time t, for t1 3< < , atwhich the velocity of particle R is 2.

(d) At time t 3= , the velocity of particle R described in part (c) is 2− . Are particles Q and R moving towardeach other or away from each other at time =t 3 ? Explain your reasoning.



END OF PART A

IF YOU FINISH BEFORE TIME IS CALLED, YOU MAY CHECK YOUR WORK ON PART A ONLY.

DO NOT GO ON TO PART B UNTIL YOU ARE TOLD TO DO SO.



CALCULUS AB SECTION II, Part B

Time—60 minutes Number of problems—4

NO CALCULATOR IS ALLOWED FOR THESE PROBLEMS.

DO NOT BEGIN PART B UNTIL YOU ARE TOLD TO DO SO.


333 33 33 3 3

3NO CALCULATOR ALLOWED

Do not w


not w

rite

beyo

nd th

is b

orde

r.

3. A continuous function g is defined on the closed interval £ £− x8 6. The graph of g, shown above, consists

of three line segments and a quarter of a circle centered at the point ( )0, 2 . Let f be the function given by

f x g t dtx

8( ) = ( )∫− .

(a) Find all values of x in the interval x8 6< <− at which f has a critical point. Classify each critical pointas the location of a local minimum, a local maximum, or neither. Justify your answers.


333 33 33 3

3 3


NO CALCULATOR ALLOWED

(c) Findf x

x xlim

4x 4 2→

( )+−

.

(d) Let h be the function defined by h xg x

x 12( ) =( )+

. Find ¢( )h 1 .

Do not w


not w

rite

beyo

nd th

is b

orde

r.

(b) Find f 0 .( )


444 44 44 4 4


Do not w


Do

not w

rite

beyo

nd th

is b

orde

r.

4. Consider the differential equationdy

dxy x2 12= ( )( + )− .

(a) Find y g x= ( ), the particular solution to the given differential equation with initial condition g 0 5( ) = .

(b) For the particular solution y g x= ( ) found in part (a), find∞

g xlimx

( )−→

.


444 44 44 4 4


4


Do

not w

rite

beyo

nd th

is b

orde

r. Do not w


(c) Let = ( )y f x be the particular solution to the given differential equation with initial condition f 1( ) =

Find the value ofd y

dx

2

2 at the point 1, 3( ). Is the graph of = ( )y f x concave up or concave down at the

point 1, 3( ) ? Give a reason for your answer.

3.


555 55 55 5 55


Do not w


Do

not w

rite

beyo

nd th

is b

orde

r.

5. The function f is defined by

( ) = + £+ >

⎧⎨⎪⎪

⎩⎪⎪f x x x x

e x

3 2 for 0

2 for 0.x

2

2

(a) Is f continuous at x 0= ? Justify your answer.

(b) Find ¢( )−f 2 and f 3¢( ).


555 55 55 5 5


5


Do not w


not w

rite

beyo

nd th

is b

orde

r.

(c) Explain why f 0¢( ) does not exist.

(d) Let g be the function given by ∫g x f t dtx

1( ) = ( )

−. Find g 1( ).


666 66 66 6 6


Do

not w

rite

beyo

nd th

is b

orde

r. Do not w


6. A hive contains 35 hundred bees at time t 0= . During the time interval £ £t0 4 hours, bees enter the hive at

a rate modeled by E t t t16 3 2( ) = − , where ( )E t is measured in hundreds of bees per hour. During the same

time interval, bees leave the hive at a rate modeled by L t t2 15( ) = +− , where ( )L t is measured in hundreds of

bees per hour.

(a) How many bees leave the hive during the time interval t0 2£ £ ?

(b) Write an expression involving one or more integrals for the total number of bees, in hundreds, in the hiveat time t for t0 4£ £ . Find the total number of bees in the hive at t 4= .


666 6 66 66 6 6



Do

not w

rite

beyo

nd th

is b

orde

r. Do not w


(c) Find the minimum number of bees in the hive for £ £t0 4. Justify your answer.


STOP

END OF EXAM


Notes on the AP Calculus AB Practice Exam

Introduction This section describes how the questions in the AP Practice Exam correspond to the components of the curriculum framework included in the AP® Calculus AB and AP® Calculus BC Course and Exam Description. For each of the questions in the AP Practice Exam, the targeted learning objectives, essential knowledge statements, and Mathematical Practices for AP Calculus from the curriculum framework are indicated.

Exam questions assess the learning objectives detailed in the curriculum framework; as such, they require a strong conceptual understanding of calculus in conjunction with the application of one or more of the Mathematical Practices for AP Calculus (MPACs). Although topics in subject areas such as algebra, geometry, and precalculus are not explicitly assessed, students must have mastered the relevant preparatory material in order to apply calculus techniques successfully and accurately.

Students take either the AP Calculus AB Exam or the AP Calculus BC Exam. The exams, which are identical in format, consist of a multiple-choice section and a free-response section, as shown in the tables on the following page.

In this publication, the multiple-choice and free-response questions include the following features:

• For multiple-choice questions, the correct response is indicated with a rationale for why it is correct. There are also explanations for the incorrect responses. Note that in the cases where multiple learning objectives, essential knowledge statements, or MPACs are provided, the primary one is listed first.

• Free-response questions include scoring guidelines that explain how students can use required knowledge learned in the AP Calculus course to answer the questions.

Student performance on these two sections will be compiled and weighted to determine an AP Exam score. Each section of the exam counts toward 50 percent of the student’s score. Points are not deducted for incorrect answers or unanswered questions.


Section I: Multiple Choice

PartGraphing Calculator

Number of Questions Time

Percentage of Total Exam Score

Part A Not permitted 30 60 minutes

50%Part B Required 15 45 minutes

TOTAL 451 hour, 45 minutes

Section II: Free Response

PartGraphing Calculator

Number of Questions Time

Percentage of Total Exam Score

Part A Required 2 30 minutes

50%Part B Not permitted 4 60 minutes

TOTAL 61 hour, 30 minutes

Calculator Use on the ExamsBoth the multiple-choice and free-response sections of the AP Calculus Exams include problems that require the use of a graphing calculator. A graphing calculator appropriate for use on the exams is expected to have the built-in capability to do the following:

• Plot the graph of a function within an arbitrary viewing window• Find the zeros of functions (solve equations numerically)• Numerically calculate the derivative of a function• Numerically calculate the value of a definite integral

One or more of these capabilities should provide sufficient computational tools for successful development of a solution to any AP Calculus AB or Calculus BC Exam question that requires the use of a calculator. Care is taken to ensure that the exam questions do not favor students who use graphing calculators with more extensive built-in features.

Students are expected to bring a graphing calculator with the capabilities listed above to the exams. AP teachers should check their own students’ calculators to ensure that the required conditions are met. Students and teachers should keep their calculators updated with the latest available operating systems. Information is available on calculator company websites. A list of acceptable calculators can be found at the AP student website (https://apstudent.collegeboard.org/apcourse/ap-calculus-ab/calculator-policy).

Note that requirements regarding calculator use help ensure that all students have sufficient computational tools for the AP Calculus Exams. Exam restrictions should not be interpreted as restrictions on classroom activities.




Section I: Multiple Choice The multiple-choice section on each exam is designed for broad coverage of the course content. Multiple-choice questions are discrete, as opposed to appearing in question sets, and the questions do not appear in the order in which topics are addressed in the curriculum framework. There are 30 multiple-choice questions in Part A and 15 multiple-choice questions in Part B; students may use a graphing calculator only for Part B. Each part of the multiple-choice section is timed and students may not return to questions in Part A of the multiple-choice section once they have begun Part B.

Curriculum Framework Alignment and Rationales for Answers

Question 1

Learning Objective Essential Knowledge Mathematical Practices for AP Calculus

3.3B(b) Evaluate definite integrals.

3.3B2: If f is continuous on the interval a b,[ ] and F is an antiderivative of f , then

MPAC 3: Implementing algebraic/computational processes

MPAC 5: Building notational fluency

(A) This option is incorrect. Lack of parentheses is a common mistake, resulting in sign errors, when the antiderivative consists of more than one term:

This answer might also be obtained if the student uses parentheses correctly but makes an algebraic mistake by taking −( ) = −1 12 to get:

(B) This option is correct. This question involves using the basic power rule for antidifferentiation and then correctly substituting the endpoints and evaluating:

(C) This option is incorrect. If the process of antidifferentiation and differentiation are confused in evaluating the definite integral, the result is:

f x dx F b F aa

b( ) = ( ) − ( )∫ .

x x3

2

1

3

3273 9 1

3 1 0 13 1 2

3− = − − −( ) − = + − = −−

.

x x3

2

1

3

3273 9 1

3 1 0 13 1 2

3− = −( ) − − − −( )( ) = + −( ) = −−

.

x x3

2

1

3

3273 9 1

3 1 43− = −( ) − − −( ) =

−

.

2 2 813x − =− .


(D) This option is incorrect. A possible error in applying the power rule for antiderivatives is to not divide by the new exponent. This error would give:

Question 2


2.1C Calculate derivatives. 2.1C3: Sums, differences, products, and quotients of functions can be differentiated using derivative rules.



(A) This option is incorrect. Subtraction was used in the product rule instead of addition.

(B) This option is incorrect. The correct product rule was used but the chain rule is ignored in the derivative of e x2 , thereby a factor of 2 was lost.

(C) This option is incorrect. The chain rule was correctly used but the product

rule is incorrectly stated as being ddx f x g x f x g x( ) ( )( ) = ′( ) ′( ).

(D) This option is correct. Use a combination of the product rule and the chain rule to compute the derivative, then evaluate that derivative at

x x3 21

32 12− =

−.

′( ) = +( ) − ( )f x e x e xx x2 1 32 3 2 2

′( ) = − =f e e e2 18 12 64 4 4

′( ) = +( ) + ( )f x e x e xx x2 3 2 21 3

′( ) = + =f e e e2 9 12 214 4 4

′( ) = ( )′( ) = ⋅( ) =f x e x

f e e

x2 3

2 3 2 242

2 2

2 44

x = 2.

′( ) = +( ) + ( )f x e x e xx x2 1 32 3 2 2

′( ) = + =f e e e2 18 12 304 4 4


Question 3


2.3B Solve problems involving the slope of a tangent line.

2.3B2: The tangent line is the graph of a locally linear approximation of the function near the point of tangency.

MPAC 2: Connecting concepts


(A) This option is incorrect. An incorrect equation was used, namely y f a f a x= ( ) + ′( ) , in constructing the equation of a line given a slope

and a point on the line.

(B) This option is correct. An equation of the tangent line at x a= is y f a f a x a= ( ) + ′( ) −( ). In this question a = 2 and ′( ) = −f a 1

2 . The value of y when x = 2 1. would be an approximation to f 2 1. .( )

(C) This option is incorrect. An error was made in constructing the equation of a line given a slope and a point on the line, using y f a f a x a= ( ) − ′( ) −( ) for the equation of the tangent line.

The same answer could also be obtained with the correct tangent line equation by making a multiplication error or sign error in substituting the derivative value or performing the calculation.

(D) This option is incorrect. This is a fundamental conceptual error in not recognizing the derivative as the slope of the tangent line.

f f2 1 2 12 2 1 4 1 05 2 95. . . .( ) ≈ ( ) − ( ) = − =

f f2 1 2 12 2 1 2 4 1

2 0 1 4 0 05 3 95. . . . .( ) ≈ ( ) − −( ) = − ( ) = − =

f f2 1 2 12 2 1 2 4 1

2 0 1 4 05. . . .( ) ≈ ( ) + −( ) = + ( ) =

f f2 1 2 2 1 2 4 0 1 4 1. . . .( ) ≈ ( ) + −( ) = + =


Question 4


2.2A Use derivatives to analyze properties of a function.

2.2A1: First and second derivatives of a function can provide information about the function and its graph including intervals of increase or decrease, local (relative) and global (absolute) extrema, intervals of upward or downward concavity, and points of inflection.



(A) This option is incorrect. The error comes from believing that g is always increasing because the derivative is the sum of two terms and is thus always positive.

(B) This option is correct. First find the critical values where the derivative is 0 or undefined. A relative extrema occurs at a critical value if the derivative changes sign there.

x x2 4 12 0+( ) = at x = 0 or x = −3There are two zeros, but ′( )g x changes sign for x = −3 only. The sign

of ′( )g x at x = 0 does not change because of the x2 factor.

(C) This option is incorrect. The number of critical values has been confused with the number of relative extrema. There are two zeros for ′( ) =g x 0, but the derivative only changes sign at one of them.

(D) This option is incorrect. The error is the result of assuming that a quartic polynomial always has 3 relative extrema or that a cubic polynomial always has 3 zeros.

′( ) = + = +( )g x x x x x4 12 4 123 2 2


Question 5


3.4B Apply definite integrals to problems involving the average value of a function.

3.4B1: The average value of a function f over an interval a b,[ ] is

MPAC 1: Reasoning with definitions and theorems


(A) This option is incorrect. The average velocity has been confused with the average rate of change of velocity over the interval:

(B) This option is incorrect. The average velocity is not the arithmetic mean of the velocity values at the endpoints:

(C) This option is correct. The average value of v t( ) over the interval 1 3,[ ] is computed with a definite integral:

(D) This option is incorrect. Speed and velocity have been confused and it was assumed that the average velocity must be positive:

1b a f x dx

a

b

−( )∫ .

v v3 13 1

7 12 4( ) − ( )

−= − − = − .

v v3 12

7 12 3( ) + ( )

= − + = − .

13 1

13 1 2 1

2 2 131

3 21

3 3

1

3

−( ) =

−−( ) = −

∫ ∫v t dt t dt t t

= −( ) − −( )( ) = −12 6 9 2 1

373 .

13 1 2 7

32

1

3

−−( ) =∫ t dt .


Question 6


3.2B Approximate a definite integral.

3.2B2: Definite integrals can be approximated using a left Riemann sum, a right Riemann sum, a midpoint Riemann sum, or a trapezoidal sum; approximations can be computed using either uniform or nonuniform partitions.



(A) This option is correct. The trapezoidal sum is the average of the left and right Riemann sums. The three intervals are of length 2, 5, and 2, respectively. Taking the average of the left and right endpoint values on each interval and multiplying by the length of the interval gives the following trapezoidal sum:

(B) This option is incorrect. Using the trapezoidal rule is not appropriate because the intervals are of different lengths. Ignoring the values of

t and believing that ∆t = − =9 03 3 as if the intervals were of equal

length in the table would yield the following computation using the trapezoidal rule:

(C) This option is incorrect. A trapezoidal sum has been confused with a left Riemann sum:

(D) This option is incorrect. The error is due to taking just the sum of the left and right Riemann sums rather than the average, thus not dividing each width by 2:

22 15 9 5

2 9 5 22 5 4 68+( ) + +( ) + +( ) = .

32 15 2 9 2 5 4 70 5+ ( ) + ( ) +[ ] = . .

2 15 5 9 2 5 85⋅ + ⋅ + ⋅ = .

2 15 9 5 9 5 2 5 4 136+ + + + +( ) ( ) ( ) = .


Question 7


3.4D Apply definite integrals to problems involving area, volume, (BC) and length of a curve.

3.4D1: Areas of certain regions in the plane can be calculated with definite integrals. (BC) Areas bounded by polar curves can be calculated with definite integrals.


MPAC 3: Implementing algebraic/computational processes3.3B(b) Evaluate definite

integrals.3.3B2: If f is continuous on the interval a b,[ ] and F is an antiderivative of f , then

f x dx F b F aa

b( ) = ( ) − ( )∫ .

(A) This option is incorrect. If a sketch is not made or if it is not checked

first whether the two graphs intersect on the interval 1 3≤ ≤x , it is likely that the calculation for area will be done with just one integral:

f x g x dx x x dx

x x

( )

.

− ( )( ) = −( )= −( ) = −( ) − −( )

∫ ∫1

3 21

3

3 21

3

6 12

2 6 54 54 2 6

If the integration had been done as g x f x dx( ) ,− ( )( )∫13

the answer

would be −4. But since area must be positive, the absolute value would give the same answer as the first integral.

(B) This option is correct. It is helpful to sketch the graphs of the two functions and to check if the two graphs cross inside the given interval. The first graph is a concave up parabola with zeros at x = 0 and x = 3 , and the second graph is a line passing through the origin. The sketch indicates that there are two regions between the curves on the interval 1 3≤ ≤x ,. The line intersects the parabola where 6 18 62x x x− = − , or 0 6 12 6 22= − = −( )x x x x ; therefore the top and bottom curves of the regions switch roles at x = 2. Different integrals should be used to find the total area of the regions, one for the interval 1 2≤ ≤x and one for the interval 2 3≤ ≤x . Let f x x x( ) = −6 182 and g x x( ) = −6 . The total area is:

1 2 3O x

y

−10

10

f x g x dx g x f x dx f x g x dx( ) − ( ) = ( ) − ( )( ) + ( ) − ( )( )∫ ∫ ∫1

3

1

2

2

3

= −( ) + −( )∫ ∫12 6 6 1221

2 22

3x x dx x x dx = −( ) + −( )6 2 2 62 3

1

2 3 22

3x x x x

= −( ) − −( )( ) + −( ) − −( )( ) = + =24 16 6 2 54 54 16 24 4 8 12.


(C) This option is incorrect. The error comes from finding the area bounded by the curves and the x-axis between x = 1 and x = 3. Since the region is below the axis, the area is the opposite of the integral value:

(D) This option is incorrect. The error comes from finding the area bounded only by the parabola and x-axis between x = 1 and x = 3. Because the region is below the axis, the area is the opposite of the integral value:

− ( )( ) + − ( )( )∫ ∫g x dx f x dx1

2

2

3.

1 2 3O x

y

−10

10

1 2 3O x

y

−10

10

− ( )∫ f x dx1

3.


Question 8


2.3B Solve problems involving the slope of a tangent line.

2.3B1: The derivative at a point is the slope of the line tangent to a graph at that point on the graph.



(A) This option is incorrect. The slope of the tangent line has been incorrectly set equal to g −( )1 instead of ′ −( )g 1 :

(B) This option is incorrect. Only the slope of the line was calculated through the two given points, but that was not connected to the calculation to find the value of b.

(C) This option is incorrect. The slope of the line through ( , )0 2− and ( , )3 4 was mistakenly found using the wrong difference quotient,

(D) This option is correct. Parallel lines have the same slope. The slope of the tangent line at x = −1 is therefore equal to the slope of the line through

( , )0 2− and ( , ),3 4 which is 4 2

3 0 2− −( )−

= . Therefore find b so that

g b−( ) = −( ) + −( ) =1 1 1 22

⇒ = −b 1.

∆∆xy =

12 .

′( ) = +g x x b2

′ −( ) = −( ) + = ⇒ =g b b1 2 1 12

52

′ −( ) =g 1 2.

′( ) = +g x x b2

′ −( ) = −( ) + = ⇒ =g b b1 2 1 2 4


Question 9


3.2C Calculate a definite integral using areas and properties of definite integrals.

3.2C3: The definition of the definite integral may be extended to functions with removable or jump discontinuities.



(A) This option is correct. This function has a jump discontinuity at x = 0. f x( ) = −1 for x < 0 and f x( ) = 1 for x > 0. The graph of f is shown to the right along with the regions A and B between the graph of f and the x-axis on the intervals −[ ]5 0, and 0 3, ,[ ] respectively.

Computing the definite integral in terms of areas and taking into account which region is below the axis and which is above, we get:

The definite integral over the interval −[ ]5 3, can also be written as the sum of the definite integrals over −[ ]5 0, and 0 3, ,[ ] giving:

(B) This option is incorrect. Sign errors were made in the definition of f so that the graph of f is reflected across the x-axis. This then yields:

or

(C) This option is incorrect. Rather than taking into account the signed areas, this is the total area:

or

(D) This option is incorrect. The source of the error may be not recognizing that the definition of the definite integral can be extended to functions with removable or jump discontinuities.

f x dx( ) = − ( ) + ( ) = − + = −−∫ 5

35 3 2area A area B .

f x dx dx dx( ) = − + = − + = −− −∫ ∫ ∫5

3

5

0

0

31 1 5 3 2.

f x dx( ) = ( ) − ( ) = − =−∫ 5

35 3 2area A area B

f x dx dx dx( ) = + − = − =− −∫ ∫ ∫5

3

5

0

0

31 1 5 3 2.

f x dx( ) = ( ) + ( ) = + =−∫ 5

35 3 8area A area B

f x dx dx dx( ) = − + = + =− −∫ ∫ ∫5

3

5

0

0

31 1 5 3 8.

x

y

1

−1

−5

3

A

B


Question 10


3.3B(b) Evaluate definite integrals. 3.3B5: Techniques for finding antiderivatives include algebraic manipulation such as long division and completing the square, substitution of variables, (BC) integration by parts, and nonrepeating linear partial fractions.



(A) This option is incorrect. The technique of substitution of variables was not applied correctly because only the substitution u x= −2 1 was made. There is no substitution for dx or for the limits of integration.

(B) This option is incorrect. The technique of substitution of variables was not applied correctly because only the substitutions for 2 1x − and dx were made, but not for the limits of integration.

(C) This option is incorrect. The technique of substitution of variables was not applied correctly because only the substitutions for 2 1x − and the limits of integration were made, but not for dx.

(D) This option is correct. Starting with the substitution u x= −2 1,

Also change the limits of integration for x to limits of integration for u:

Substituting for 2 1x − , for dx, and for the limits of integration gives:

u x du dx dx du= − ⇒ = ⇒ =2 1 2 12 .

x u= ⇒ = ⋅ − =2 2 2 1 3x u= ⇒ = ⋅ − =3 2 3 1 5.

g x dx g u du g u du2 1 12

122

3

3

5

3

5−( ) = =( ) ⋅ ( )∫ ∫ ∫ .


Question 11



3.2C1: In some cases, a definite integral can be evaluated by using geometry and the connection between the definite integral and area.


MPAC 4: Connecting multiple representations

(A) This option is incorrect. Only the area of the semicircle was considered. The opposite of that area was then taken, since the region is below the axis.

(B) This option is incorrect. Only the area of the semicircle was considered, ignoring both the area of the rectangle and the fact that the region is below the axis.

(C) This option is correct. The definite integral can be evaluated by using geometry and the connection of the definite integral with areas. In particular, the value of the integral here is the opposite of the area of the region between the semicircle and the x-axis, i.e. the area of the rectangle with vertices at 2 0, ,( ) 12 0, ,( ) 2 6, ,−( ) and 12 6, −( ) minus the area of the semicircle.

The area of the rectangle is 6 10 60⋅ = . The area of the semicircle of

radius 5 is 12 5 25

22π π( ) = .

The definite integral is therefore equal to

− −( )60 252π .

(D) This option is incorrect. Only the area of the rectangle minus the area of the semicircle was computed; the need to take the opposite of this value, since the region is below the axis, was not recognized.

(2, −6) (12, −6)


Question 12


3.5A Analyze differential equations to obtain general and specific solutions.

3.5A1: Antidifferentiation can be used to find specific solutions to differential equations with given initial conditions, including applications to motion along a line, exponential growth and decay, (BC) and logistic growth.



(A) This option is incorrect. This is the velocity at t = 2, not the position:

v t t v( ) = + ⇒ ( ) =3 10 2 222 .

The velocity at t = 2 can also be found directly using the Fundamental Theorem of Calculus:

(B) This option is incorrect. The assumption was made that the velocity is constant and the acceleration was ignored:

(C) This option is incorrect. This is the total change in position without accounting for the initial position. The calculation uses the correct velocity function but takes p t t t( ) = +3 10 .

(D) This option is correct. Velocity is the antiderivative of acceleration and position is the antiderivative of velocity. In each case, the object’s velocity and position at t = 0 can be used to find the appropriate “+ C” after each antidifferentiation. The last step is to then evaluate the position function at t = 2.

v v v t dt a t dt t dt2 0 10 10 6 220

2

0

2

0

2( ) = ( ) + ′( ) = + ( ) = + =∫ ∫∫ .

p v2 7 2 0 7 2 10 27( ) = + ⋅ ( ) = + ⋅ = .

a t t( ) = 6

v t t C

v C C

v t t

( ) = +

( ) = ( ) + = ⇒ =

( ) = +

3

0 3 0 10 10

3 10

21

21 1

2

p t t t C

p C C

p t t t

( ) = + +

( ) = ( ) + ( ) + = ⇒ =

( ) = + +

32

32 2

3

10

0 0 10 0 7 7

10 7

p 2 2 10 2 7 353( ) = + ( ) + =


Question 13


2.1D Determine higher order derivatives.

2.1D1: Differentiating f ′ produces the second

derivative f ′′, provided the derivative of f ′ exists; repeating this process produces higher order derivatives of f.



(A) This option is correct. This question requires repeated differentiation using knowledge of the derivatives of sine and cosine, the derivative of the natural logarithm, and the power rule for derivatives, as well as recognizing the need to use the chain rule for the first derivative.

(B) This option is incorrect. A consistent sign error was made with the derivatives of the two trigonometric functions by using

ddx x xsin cos( ) = − and

ddx x xcos sin .( ) =

dydx x x x xd ydx

xx

d ydx

xx

= − − = − −

= − +

= −

sin sin

cos

sin

22

1

1

2

2

2 2

3

3 3

dydx x x x xd ydx

xx

d ydx

xx

= − = −

= − +

= − −

sin sin

cos

sin

22

1

1

2

2

2 2

3

3 3


(C) This option is incorrect. The only error was not using the chain rule when taking the first derivative. No chain rule is needed after that:

(D) This option is incorrect. A chain rule error was made on the first derivative, and there are consistent sign errors with the derivatives of the trigonometric functions.

dydx x xd ydx

xx

= − −

= − +

sin

cos

12

12

2

2 2

d ydx

xx

xx

3

3 3 32

21= − = −sin sin .

dydx x xd ydx

xx

d ydx

xx

xx

= −

= − +

= − − = − −

sin

cos

sin sin

12

122

21

2

2 2

3

3 3 3


Question 14


3.3A Analyze functions defined by an integral.

3.3A3: Graphical, numerical, analytical, and verbal representations of a function f provide information about the function g defined as





(A) This option is incorrect. The Fundamental Theorem of Calculus was applied to calculate values of g x( ), but all the areas were added to the value g 2( ) . Therefore the conclusion was that g x( ) ≥ 5 for all values of x.

(B) This option is incorrect. It comes from confusing the graph of g with the graph of f and finding the number of times that f x( ) = 3, which happens only between x = 5 and x = 6.

(C) This option is incorrect. The Fundamental Theorem of Calculus was applied correctly to find values of g x( ) for x > 2, thus finding the

solutions c = 4 and 5 6< <c . However, the integral f x dx( )∫20

might

have been mishandled, thus missing the solution at c = 0. (D) This option is correct. By the Fundamental Theorem of Calculus and using

area and geometry to calculate the definite integral,

g g g x dx f x dx0 2 5 5 2 32

0

0

2( ) = ( ) + ′( ) = − ( ) = − =∫ ∫ .

So c = 0 is one solution to g x( ) = 3. Since ′ =g f , the graph of f indicates that g is increasing from x = 0 to x = 2 since f is positive there. From x = 2 to x = 4 the function g decreases by the same amount that it increased from x = 0 to x = 2 by the symmetry of the regions.

g f x dx4 5 5 2 32

4( ) = + ( ) = + −( ) =∫

This gives a second solution at c = 4. According to the graph of the derivative f, the function g continues to decrease until x = 5 and then begins increasing again.

Since g is continuous, the Intermediate Value Theorem guarantees a third solution somewhere on the interval ( , ).5 6 There are no other solutions possible. (The third solution is at c = 5 5. . )

g x f t dta

x( ) = ( )∫ .

g f x dx

g f x dx

5 5 5 3 2

6 2 2 2 4

2

5

5

6

( ) = + = + −( ) =

( ) = + = + =

∫∫

( )

( )


Question 15


1.2A Analyze functions for intervals of continuity or points of discontinuity.

1.2A1: A function f is continuous at x c= provided that f c( ) exists, limx c

f x→

( ) exists, and

lim .x c

f x f c→

( ) = ( )



(A) This option is incorrect. Rather than setting as equal the left- and right-sided limits at x = 1, the derivatives of the two pieces were set equal at x = 1.

(B) This option is correct. Since the function f is continuous at x = 1, then lim lim .x x

f x f x→ →− +

( ) = ( )1 1

Evaluate these two limits and set them equal,

then solve for c.

(C) This option is incorrect. Rather than setting as equal the left- and right-sided limits at x = 1, the derivatives of the two pieces were set equal at x = 1, but a calculus error was made with respect to the derivative of a constant.

ORThe left- and right-sided limits were set equal at x = 1, but an algebraic error was made in evaluating ln :1 1=

(D) This option is incorrect. The value of the function at x = 1 was confused with c so that c f= ( ) = + =1 9 2 1 9ln .

ddx cx d

dx x

c x c

6 9 2

2 2

+( ) = +( )

= ⇒ =

ln

limx

f x c c→ −

( ) = + ⋅ = +1

6 1 6

lim lnx

f x→ +

( ) = + =1

9 2 1 9

6 9 3+ = ⇒ =c c

ddx cx d

dx x

c xc c

6 9 2

6 9 2

6 9 2 5

+( ) = +( )

+ = +

+ = + ⇒ =

ln

6 1 9 2 1 11 5+ ( ) = + ( ) = ⇒ =c c .


Question 16


2.3F Estimate solutions to differential equations.

2.3F1: Slope fields provide visual clues to the behavior of solutions to first order differential equations.



(A) This option is incorrect. For this differential equation, the slopes of the line segments in Quadrant II must be positive since y > 0 and x2 0> in that quadrant. In this slope field that does not happen, as can be observed with the segments near the bottom left of Quadrant II. In addition, all line segments along the x-axis should have positive slopes.This is not the case here.

This slope field might be chosen if the squared term is not accounted for.

[This is the slope field for dydx x y= + . ]

(B) This option is incorrect. For this differential equation, the slopes of the line segments in Quadrant I must be positive since y > 0 and x2 0> in that quadrant. In this slope field, however, all the line segments in Quadrant I have negative slopes. In addition, all line segments along the x-axis should have positive slopes. This is not the case here.

This slope field might be chosen if one considers dydx = 0

and thinks of

the differential equation as relating to the parabola y x= − 2.

[This is the slope field for dydx x= −1 8. . ]

(C) This option is incorrect. For this differential equation, the slopes of the line segments in Quadrant II must be positive since y > 0 and x2 0> in that quadrant. In this slope field, however, that does not happen as can be observed with the segments near the bottom left of Quadrant II. In addition, all line segments along the x-axis should have positive slopes. This is not the case here.

This option might be chosen if the x and y variables are confused and

one looks for the slope field for

dydx x y= + 2.

(D) This option is correct. The line segments in the slope field have slopes

given by dydx x y= +2 at the point x y, .( ) In Quadrants I and II, all

slopes must be positive or zero since y > 0 in those quadrants and

x2 0≥ . This is the only option in which that condition is true.


Question 17


2.3E Verify solutions to differential equations.

2.3E2: Derivatives can be used to verify that a function is a solution to a given differential equation.



(A) This option is correct. One way to verify that a function is a solution to a differential equation is to check that the function and its derivatives satisfy the differential equation. The differential equations in this problem involve both ′y and ′′y .. The correct derivatives must be computed and the algebra correctly done to verify that the differential equation is satisfied.

(B) This option is incorrect. The correct derivatives were found but this differential equation may appear to be satisfied if the subtraction in the second term is not distributed correctly across the two terms inside the parentheses:

(C) This option is incorrect. This differential equation will appear to be satisfied if the chain rule is not used in taking the derivative of the exponential:

(D) This option is incorrect. This differential equation will appear to be satisfied if the chain rule is not used in taking the derivative of the exponential and the subtraction in the second term is not distributed correctly across the two terms inside the parentheses:

′ = −

′′ =

′′ − ′ − = − −( ) −= − +

y ey ey y e e

e e

x

x

x x

x x

3 593 15 9 3 3 5 15

9 9 15

3

3

3 3

3 3 −− =15 0

′′ − ′ + = − −( ) + = − − + =y y e e e ex x x x3 15 9 3 3 5 15 9 9 15 15 03 3 3 3 .

′ = −

′′ =

y ey e

x

x

3

35

(error)

′′ − ′ − = − −( ) − = − + − =y y e e e ex x x x5 5 5 5 5 03 3 3 3 .

′ = −

′′ =

y ey e

x

x

3

35

(error)

′′ − ′ + = − −( ) + = − − + =y y e e e ex x x x5 5 5 5 5 03 3 3 3 .


Question 18


2.1C Calculate derivatives. 2.1C2: Specific rules can be used to calculate derivatives for classes of functions, including polynomial, rational, power, exponential, logarithmic, trigonometric, and inverse trigonometric.



(A) This option is incorrect. The incorrect derivative, ddx x xsin cos ,− −=1 1

was used but the inverse cosine function was correctly evaluated to get

cos .−

=

1 32 6

π

(B) This option is incorrect. The incorrect derivative, ddx x xsin cos− −=1 1,

was used and the inverse cosine function was incorrectly evaluated as

cos .−

=

1 32 3

π

(C) This option is incorrect. The derivative of the inverse sine function was confused with the derivative of the inverse tangent function, and

therefore 11 2+ x

was used for the derivative:

(D) This option is correct. The derivative of the inverse sine function sin−1 x

is

11 2− x

. Since

1

1 32

1

1 34

472

+

=+

= .

f x x( ) = −sin ,1

′

=

−

= =f 32

1

1 32

114

22

.


Question 19


2.1A Identify the derivative of a function as the limit of a difference quotient.

2.1A2: The instantaneous rate of change of a function at a point can be expressed

by limh

f a h f ah→

+( ) − ( )0

or lim ,x a

f x f ax a→

( ) − ( )−

provided that the limit exists. These are common forms of the definition of the derivative and are denoted f a′( ).



(A) This option is incorrect. This could have been caused by observing that the numerator is zero at x e= and then ignoring the denominator.

(B) This option is correct. The limit of this difference quotient is of the form

limx a

f x f ax a→

( ) − ( )−

where f x x x( ) = −20 3 and a e= . This is one way to

express the derivative of f at x e= .

(C) This option is incorrect. This is just the value of the function x x20 3− at x e= .

(D) This option is incorrect. Observing that the denominator was zero at x e= could have led to the assumption of unbounded behavior while the numerator was ignored. Or it was thought that the indeterminate

form 00 produced a nonexistent result, without recognizing the limit as

involving the definition of the derivative.

′( ) = −( ) = −

⇒ ′( ) = −

f x ddx x x x

f e e

20 19

19

3 20 3

20 3


Question 20


2.1C Calculate derivatives. 2.1C5: The chain rule is the basis for implicit differentiation.




2.1D1: Differentiating f ′ produces the second

derivative f ′′, provided the derivative of f ′ exists; repeating this process produces higher order derivatives of f.

(A) This option is incorrect. The chain rule may not have been used.

The x and y values could also have been reversed when substituting into the second derivative.

(B) This option is incorrect. There was an error in the power rule when differentiating y3 resulted in 3y.

(C) This option is correct. The answer is obtained by using the chain rule when doing implicit differentiation. The second derivative is found in

terms of y and dydx , which can then be written completely in terms of y.

The condition f 1 2( ) = means that y = 2 when x = 1.

d ydx

y d ydx x y

2

22

2

21 2

3 3 4 12= ⇒ = ( ) =( )=( ), ,

d ydx

y dydx y y

d ydx x y

2

22 2 3

2

21 2

3 3 3

3 1 4 12

= = +( )

= ( )( ) =( )=( ), ,

d ydx

y dydx y y

d ydx x y

2

23

2

21 2

3

3 3 3

3 2 2 3 3 2 11

= = +( )

= ( ) +( ) = ( )(( )=( ), ,

)) = 66

d ydx

y dydx y y

d ydx x y

2

22 2 3

2

21 2

2 3

3 3 3

3 2 2 3 3 4 1

= = +( )

= ⋅ +( ) = ( )( )=( ), ,

11 132( ) =


(D) This option is incorrect. There was an error in differentiating a constant function and not getting 0 (i.e. not differentiating the 3).

Question 21


2.1C Calculate derivatives. 2.1C4: The chain rule provides a way to differentiate composite functions.



(A) This option is incorrect. The notation f g x( )( ) was confused with the product f x g x( ) ( ) and the product rule was used to find the derivative:

(B) This option is correct. The solution is found by using the chain rule correctly and then using values of the functions and their derivatives from the table to evaluate the derivative at x = 1.

(C) This option is incorrect. The chain rule for composite functions was

incompletely understood by thinking that ddx f g x f g x( )( ) = ′ ( )( ) :

(D) This option is incorrect. The chain rule for composite functions was

misunderstood by thinking that ddx f g x f g x( )( ) = ′ ′( )( ) :

d ydx

y dydx y y

d ydx x y

2

22 2 3

2

21 2

3 3 3 3 3

3 4 3 1

= +( ) = +( ) +( )

= ⋅ +( )( )=( ), ,

11 165( ) =

′( ) = ′( ) ( ) + ′( ) ( )h f g g f1 1 1 1 1′( ) = −( )( ) + −( )( ) = −h 1 3 3 2 5 19.

′( ) = ′ ( )( ) ′( )′( ) = ′( ) ′( )′( ) = −( ) = −

h f g gh f gh

1 1 11 3 11 7 2 14

′( ) = ′ ( )( ) =h f g1 1 7.

′( ) = ′ ′( )( )′( ) = ′ −( )h f gh f

1 11 2

′( ) =h 1 9.


Question 22



3.5A2: Some differential equations can be solved by separation of variables.



(A) This option is incorrect. The arbitrary constant must be included when doing the antidifferentiation, not after solving for y (sometimes referred to as “The Late C”). This is the result if the constant is introduced at the wrong time.

(B) This option is incorrect. The arbitrary constant was introduced after solving for y (see the steps in option (A)), and the positive square root was selected.

(C) This option is correct. This question involves solving a differential equation by separation of variables and using the initial condition to determine the appropriate value for the arbitrary constant. The solution leads to y2. Since the initial value of y is negative, in solving for y the negative square root is used.

(D) This option is incorrect. All the calculus steps were done correctly as in the calculations for option (C), but the positive square root was used at the last step without consideration of the sign of the initial y value.

dydx

xy y dy x dx

y dy x dx y x x y x x

y x

= + ⇒ = +( )

= +( ) ⇒ = + ⇒ = +

= ±

∫ ∫

1 1

1 2 2 22 2

2 2

2 ++ +− = +

= − + −

22 0

2 22

x CC

y x x

dydx

xy y dy x dx

y dy x dx y x x C

C

= + ⇒ = +( )

= +( ) ⇒ = + +

−( )= + +

∫ ∫

1 1

1 2 222 0 0

2 2

2⇒⇒ =

= + + ⇒ = + +

= − + +

2

2 2 2 2 4

2 4

2 22 2

2

C

y x x y x x

y x x


Question 23


3.4D Apply definite integrals to problems involving area, volume, (BC) and length of a curve.

3.4D2: Volumes of solids with known cross sections, including discs and washers, can be calculated with definite integrals.



(A) This option is incorrect. This expression uses the entire region between the curves from x = 0 to x = 4, including the triangle below the x-axis from x = 0 to x = 2 that is not part of region R. It treats every rotated slice on the entire interval as if forming a washer.

(B) This option is incorrect. This expression uses the entire region between the curves from x = 0 to x = 4, including the triangle below the x-axis from x = 0 to x = 2 that is not part of region R. It treats every rotated slice on the entire interval as forming a washer, and also makes the mistake of using π r r2 1

2−( )

instead of π r r22

12−( ) for the area of a

washer with inner radius r1 and outer radius r2.

(C) This option is correct. The region R must be split into two parts at x = 2 when considering its rotation around the x-axis. For the left part, rotating a typical slice around the x-axis will form a disc of radius r x= . For the right part, rotating a typical slice around the x-axis will form a washer with inner radius r x1 2= − and

outer radius r x2 = . The combinedvolume will be given by the integral expression:

(D) This option is incorrect. The region was correctly split into two parts at x = 2, but the second integral used π r r2 1

2−( )

instead of π r r22

12−( )

for the area of a washer with inner radius r1 and outer radius r2.

π πx dx x x dx( ) + ( ) − −( )( )∫ ∫2

0

2 2 22

42

2x

y

4O

= + − −( )( )∫ ∫π πx dx x x dx0

2 22

42 .


Question 24


1.1C Determine limits of functions.

1.1C3: Limits of the

indeterminate forms 00

and ∞∞ may be evaluated

using L’Hospital’s Rule.



(A) This option is incorrect. It may have been observed that the numerator is zero at x = 3 and the denominator was then ignored.

(B) This option is incorrect. The indeterminate form 00

was recognized and

an attempt was made to use L’Hospital’s Rule, but an error was made in the derivative of the denominator:

(C) This option is correct. Substituting x = 3 into the fraction yields the

indeterminate form 00 .

Therefore L’Hospital’s Rule can be used to find

the limit:

(D) This option is incorrect. The observation that the denominator is zero at x = 3 led to the assumption of unbounded behavior, while the numerator was ignored.

lim tan lim sec .x x x x

xe x

xe→ − → −

−( )−

=−( )

=3 3 3

2

33

33

313

lim tan lim sec .x x x x

xe x

xe→ − → −

−( )−

=−( )−

=3 3 3

2

33

33

3 112


Question 25


2.4A Apply the Mean Value Theorem to describe the behavior of a function over an interval.

2.4A1: If a function f is continuous over the interval a b,[ ] and differentiable over the interval a b, ,( ) the Mean Value Theorem guarantees a point within that open interval where the instantaneous rate of change equals the average rate of change over the interval.



(A) This option is incorrect. The error comes from thinking that the Mean Value Theorem guarantees a value of x in the open interval where the instantaneous rate of change of f is equal to the total change in f instead of the average rate of change over the interval.

′( ) = ( ) − ( ) = − =f x f f3 1 11 9 2

3 6 2 3 6 2

66

2

22 2

2

xx

x x

xx

− = ⇒ − =

==

(B) This option is correct. Because f is differentiable for x > 0, it is continuous on the closed interval 1 3,[ ] and differentiable on the open interval 1 3, .( ) Therefore the Mean Value Theorem guarantees that there is a number x with 1 3< <x satisfying:

Solve this equation for the value of x in the interval 1 3< <x .

(C) This option is incorrect. This is the x-value on the interval 1 3< <x where ′( ) =f x 0. The Mean Value Theorem may have been confused with the Extreme Value Theorem.

(D) This option is incorrect. This error comes from thinking that the Mean Value Theorem guarantees a value of x in the interval that is equal to the average rate of change over the interval:

′( ) = −−

= −−

=f x f f( ) ( ) .3 13 1

11 93 1 1

3 6 1 3 6

2 63

2

22 2

2

xx

x x

xx

− = ⇒ − =

==

x f f=

( ) − ( )−

= −−

=3 13 1

11 93 1 1.


Question 26



3.3B5: Techniques for finding antiderivatives include algebraic manipulation such as long division and completing the square, substitution of variables, (BC) integration by parts, and nonrepeating linear partial fractions.

3.3B2: If f is continuous on the interval a b,[ ] and F is an antiderivative of f, then



(A) This option is correct. Before doing the integration, the integrand can be simplified using long division.

Then:

.

(B) This option is incorrect. The error comes from believing that the definite

integral f xg x dx( )( )

⌠⌡

1

2 is equal to f x dx( )∫1

2

divided by g x dx( )∫1

2.

f x dx F b F aa

b( ) = ( ) − ( )∫ .

)x x xx

x xxx

+ − −−

+− −− −

2 53

23 53 6

1

2

2

x xx dx x x dx

2

1

2

1

252 3 1

2− −+

= − ++( )⌠

⌡ ⌠

⌡

= − + +( ) = − +( ) − − +( ) = − +x x x2

1

2

2 3 2 4 4 52 3 3

243ln ln ln ln

13

12 5

12 2

256

72

2521

3 2

1

2

2

1

2

x x x

x x

− −

+=−

= −


(C) This option is incorrect. An algebraic error was made in the long division where terms other than the leading ones were added instead of being subtracted.

)x x xx

x xxx

+ − −+

+−+−

2 51

2523

2

2

This would lead to:

(D) This option is incorrect. The antiderivative of

f xg x( )( )

is not the

antiderivative of f x( ) divided by the antiderivative of g x( ).

x xx dx x x dx

2

1

2

1

252 1 2

3− −+

= + −+( )⌠

⌡ ⌠

⌡ †

= + − +( ) = −( ) − −( ) = +x x x2

1

2

2 3 2 4 3 4 32 3 3 5

2 3 34ln ln ln ln .

13

12 5

12 2

149

3115

2345

3 2

2

1

2x x x

x x

− −

+

= − + =


Question 27


2.3C Solve problems involving related rates, optimization, rectilinear motion, (BC) and planar motion.

2.3C2: The derivative can be used to solve related rates problems, that is, finding a rate at which one quantity is changing by relating it to other quantities whose rates of change are known.



(A) This option is correct. The goal in this related rates problem is to find the

value of dAdt

at the instant when h = 3 and dhdt = − 1

2 (the derivative is

negative since the depth of the water is decreasing). Using the chain rule gives:

The area is therefore decreasing at the rate of dAdt h=

= −( ) −( ) = −3

12 6 12 3π π π . square meters per

minute.(B) This option is incorrect. A common mistake in related rates problems is

to not use the chain rule, thereby having no dhdt term to include in the

computation:

(C) This option is incorrect. Each term in A was differentiated individually, but the chain rule should have been used with the first term because there is no longer an h variable present. The fact that the depth of the water is decreasing also was not taken into account.

(D) This option is incorrect. This is the value of the area when h = 3, not

the rate of change of the area.

dAdt

dAdh

dhdt h dhdt= ⋅ = −( )12 2π π

dAdt h=

= −( ) −( ) = −3

12 6 12 3π π π .

dAdt h= −( )12 2π π

dAdt h=

= −( ) =3

12 6 6π π π .

dAdt h dhdtdAdt h

= −

= − ( )( ) ==

12 2

12 2 3 12 9

3

π π

π π π


Question 28



2.3C3: The derivative can be used to solve optimization problems, that is, finding a maximum or minimum value of a function over a given interval.



(A) This option is incorrect. This is the area of the triangle with the third

vertex at the endpoint for x = 12 :

For x = 12 ,

(B) This option is incorrect. The chain rule was not used in calculating the derivative of the area function.

(C) This option is correct. The first step is to express the area of the triangle in terms of the variable x. To find the maximum area, the next step is to find the critical value for the area function.

is the only critical value.

For x = 2, A = ⋅( ) − + = +2 2 2 2 10 2 4 8ln ln .Since there is only one critical value and it is a local maximum by the Second Derivative Test, it must be the absolute maximum.

A bh x x x x= = ( ) ( ) − +( ) = ( ) − +12

12 4 2 1

2 5 2 2 10ln ln .

A = ⋅( ) − + = − + =2 2 12

12 10 0 1

2 10 192ln .

A bh x x x x= = ( ) ( ) − +( ) = ( ) − +12

12 4 2 1

2 5 2 2 10ln ln

dAdx x x= ( ) − = −2 1

2 1 1 1

dAdx x x= − = ⇒ =1 1 0 1

A = ⋅( ) − + = +2 2 1 1 10 2 2 9ln ln

1 5

h

x 8

0.5

A bh

x x

x x

=

= ( ) ( ) − +( )= ( ) − +

12

12 4 2 1

2 5

2 2 10

ln

ln

dAdx x x x= ( ) − = − = ⇒ =2 2

2 1 2 1 0 2


(D) This option is incorrect. This is the area of the triangle with the third vertex at the endpoint for x = 8 :

For x = 8, A = ⋅( ) − + = +2 2 8 8 10 2 16 2ln ln .

Question 29


2.1C Calculate derivatives. 2.1C6: The chain rule can be used to find the derivative of an inverse function, provided the derivative of that function exists.



(A) This option is incorrect. A misconception that the tangent line for the inverse function is perpendicular to the tangent line for the function could have led to using a slope that is the “negative reciprocal”, i.e.,

believing that ddx f x f x− ( ) = − ′( )

1 1 :

(B) This option is incorrect. The notation g x f x( ) = ( )−1 may have been

confused with the reciprocal of f x( ), with the attempt then to findddx f x

1( )

by using the quotient rule (or a power rule along with the

chain rule):

For x = 2, this gives:

A bh x x x x= = ( ) ( ) − +( ) = ( ) − +12

12 4 2 1

2 5 2 2 10ln ln .

f x x x( ) = + +3 4 2

′( ) = +f x x3 42

′( ) = ⋅ + =f 2 3 2 4 162 .

ddx x x

x x x

x x

14 2

4 2 0 1 3 4

4 23

3 2

3 2+ +

=

+ +( )( ) − ( ) +( )+ +( )

.

− ⋅ +( )+ ⋅ +( )

= − = −3 2 4

2 4 2 2

1618

481

2

3 2 2 .


(C) This option is correct. Since f g x x( )( ) = , the chain rule can be used to

determine that ′ ( )( ) ′( ) =f g x g x 1. Substituting x = 2 gives

1 2 2 0 2= ′ ( )( ) ′( ) = ′( ) ′( )f g g f g .

Therefore

Alternatively, one can use the following formula that follows from the same chain rule calculation as above when starting with f f x x− ( )( ) =1 :

(D) This option is incorrect. An attempt might have been made to use the memorized formula for the derivative of an inverse function shown in the solution in option (C), but taking the reciprocal was forgotten,

leading to computing ′ ( )( )−f f 1 2 :

′( ) = ′( ) = ⋅ +=g f2 1

01

3 0 4142 .

ddx f x

f f fx

−

= −( ) =′ ( )( )

= ′( ) = ⋅ +=1

2 1 21

210

13 0 4

14 .

′( ) = +f x x3 42

′ ( )( ) = ′( ) = ⋅ + =−f f f1 22 0 3 0 4 4.


Question 30


3.2A(a) Interpret the definite integral as the limit of a Riemann sum.

3.2A3: The information in a definite integral can be translated into the limit of a related Riemann sum, and the limit of a Riemann sum can be written as a definite integral.



(A) This option is incorrect. This sum used rectangles of width ∆x n= 1 and

right endpoints corresponding to step sizes of width 1n .

These would

be the values for a right Riemann sum on the interval 2 3, .[ ] This limit is

therefore x dx22

3∫ .

(B) This option is incorrect. This sum used step sizes of width 1n

between

the right endpoints which is not consistent with the width 3n

used for

the rectangles. This limit is 3 22

3x dx∫ .

(C) This option is incorrect. This sum used step sizes of width3n

between

the right endpoints which is not consistent with the width 1n

used for

the rectangles. This limit is 13

22

5x dx∫ .

(D) This option is correct. For this integral, a right Riemann sum with n

terms is built from rectangles of width ∆x n n= − =5 2 3 . The height

of each rectangle is determined by the function x2 evaluated at theright endpoints of the n intervals. These endpoints are the x-values at 2 + k x∆ , for k from 1 to n, since the integral starts at 2. The Riemann

sum can be written as k

nk x x

=∑ +( )

1

22 ∆ ∆ where ∆x n= 3 , and its limit as

n → ∞ is x dx22

5∫ .


Question 76


3.4E Use the definite integral to solve problems in various contexts.

3.4E1: The definite integral can be used to express information about accumulation and net change in many applied contexts.



(A) This option is correct. The definite integral of the rate of change over an interval gives the net change over that interval. Since s t( ) is the rate of change at which sand is added to the beach, the definite integral of s t( ) on the interval 2 5≤ ≤t gives the net change in the amount of sand on the beach over that time period. Since t is measured in hours since 5:00 a.m., the time 7:00 a.m. corresponds to t = 2 and 10:00 a.m. corresponds to t = 5.

(B) This option is incorrect. The definite integral was correctly used to find a net change given a rate of change, but incorrectly uses the 3-hour period from t = 0 to t = 3 by not adjusting for the variable t representing the time since 5:00 a.m.

(C) This option is incorrect. This is the average value of the rate at which sand is being added over the interval 2 5≤ ≤t :

(D) This option is incorrect. This is the net change s s5 2( ) − ( ) in the rate at which sand is being added over the interval 2 5≤ ≤t , not the net change in the amount of sand on the beach over that interval.

s t dt t dt( ) = + ( )( ) =∫ ∫2

5

2

565 24 0 3 255 368sin . .

s t dt( )∫03

s t dt( )−

∫25

5 2 .


Question 77


1.1B Estimate limits of functions.

1.1B1: Numerical and graphical information can be used to estimate limits.



(A) This option is incorrect. The value a = 4 is missed, perhaps because of a mistaken belief that the function must be continuous at the point for the limit to exist.

(B) This option is correct. limx a

f x→

( ) = 0 if the left-sided and right-sided limits

at x a= both equal 0, i.e.

For this graph, that happens at x = 2 and x = 4. Note that the function does not need to be defined at x = 4 for the limit to exist there.

(C) This option is incorrect. These are the values for which f x( ) = 0 and at least a one-sided limit exists. However, lim

x af x

→( )

does not exist for

a = 0.

(D) This option is incorrect. These are the values where f x( ) = 0. However,lim ( )x a

f x→

does not exist for a = 0 and the limit is equal to −1, not 0, for

a = 1.

lim lim .x a x a

f x f x→ →− +

( ) = ( ) = 0


Question 78






(A) This option is incorrect. This might be selected if a badly chosen viewing window (i.e. not examining the correct interval) omits the first two and last two intersections.

(B) This option is incorrect. This might be selected if the viewing window omits the first two intersections or the last two intersections, or if the first two intersections are overlooked, perhaps because the range on the y-axis is set too large.

(C) This option is incorrect. This is the apparent number of points of inflection on the graph of ′′f . It is also the number of critical points of ′′f on the open interval 0 3< <x .

(D) This option is correct. Graph ′′( )f x on the interval 0 3≤ ≤x .

A point of inflection occurs where the second derivative ′′( )f x changes sign. This happens five times on the interval: twice between 0 and 1, once between 1 and 2, and twice between 2 and 3.

x

y

1 2 3

−1

1


Question 79


3.4C Apply definite integrals to problems involving motion.

3.4C1: For a particle in rectilinear motion over an interval of time, the definite integral of velocity represents the particle’s displacement over the interval of time, and the definite integral of speed represents the particle’s total distance traveled over the interval of time.



(A) This option is incorrect. This is the difference between the displacement and the total distance over the whole interval 0 5≤ ≤t .

(B) This option is correct. Let the total area above the t-axis be A and the area below the t-axis be B. The particle’s total distance on the time interval 0 5≤ ≤t . is:

and the particle’s displacement is:

Solving the system gives A = 8 and B = 5. Since the graph of v t( ) is below the axis for 2 4< <t ,

(C) This option is incorrect. v t dt B( ) = =∫2

45

was concluded by thinking

just about area and not that the definite integral corresponds to a region below the axis.

(D) This option is incorrect. This is the difference between the total distance and the displacement over the whole interval 0 5≤ ≤t .

v t dt A B( ) = + =∫05

13,

v t dt A B( ) = − =∫05

3.

v t dt B( ) = − = −∫24

5.


Question 80


3.4A Interpret the meaning of a definite integral within a problem.

3.4A2: The definite integral of the rate of change of a quantity over an interval gives the net change of that quantity over that interval.



(A) This option is incorrect. If T t( ) was the room temperature, this would

be the interpretation for the expression T H t dt00

6( ) + ( )∫

that would be

used to find the temperature at t = 6 given the initial value of 20.

(B) This option is incorrect. If T t( ) was the room temperature, this would be

an interpretation for the expression 16 0

6T t dt( )∫ .

(C) This option is correct. The definite integral of the rate of change over an interval gives the net change over that interval. Since H t( ) is the rate of change of temperature, the definite integral of H t( ) on the interval 0 6≤ ≤t gives the change in the temperature over that time period.

(D) This option is incorrect. This is an interpretation of H 6( ).


Question 81



2.2A3: Key features of the graphs of f, f ′,, and f ′′ are related to one another.



(A) This option is correct. Where the derivative ′f is positive, the function f should be increasing. Where the derivative ′f is negative, the function f should be decreasing. The derivative to the left of x = 0 is always positive. The derivative to the right of x = 0 is positive until x = 1, then negative. Therefore the graph of f should be increasing for x < 0 and 0 1< <x , and should be decreasing for x > 1. The graph in this option could therefore be the graph of f.

(B) This option is incorrect. The sign of the derivative on the left side and right side of x = 0 was misinterpreted, thus reversing the desired increasing/decreasing behavior that the graph of f should display.

(C) This option is incorrect. The sign of the derivative on the left side of x = 0 was misinterpreted, thus reversing the desired increasing behavior that the graph of f should display on the left side of x = 0.

(D) This option is incorrect. The sign of the derivative on the right side of x = 0 was misinterpreted, thus not achieving the desired increasing/decreasing behavior that the graph of f should display on the right side of x = 0.


Question 82






(A) This option is correct. The graph of ′f on the interval − < <3 3x shows

that the function f will have a relative maximum where the derivative changes from positive to negative. This happens between −2 and −1, and between 2 and 3. The calculator is used to find the zeros of ′f on those intervals.

(B) This option is incorrect. These are the values where the derivative changes from negative to positive. They would be where f has a relative minimum, not a relative maximum.

(C) This option is incorrect. These are the values where the derivative has a relative maximum, not where the function has a relative maximum.

(D) This option is incorrect. These are all of the values where the derivative is equal to zero. They are critical values, but the change in the sign of the derivative has not been used to distinguish between a relative maximum and a relative minimum.

x

y

−1−2−3 1 2 3


Question 83



1.2A3: Types of discontinuities include removable discontinuities, jump discontinuities, and discontinuities due to vertical asymptotes.



(A) This option is incorrect. This might be chosen because the function f is not differentiable at x = 1. However, it is continuous at this point.

(B) This option is incorrect. The type of discontinuity was misidentified. Since lim lim ,

x xf x f x

→ →− +( ) = ( ) =

3 30

there is no jump discontinuity at

x = 3. The function has a removable discontinuity at x = 3.(C) This option is correct. A jump discontinuity occurs at x c= if the

left- and right-sided limits exist at c, but are not equal, that is, lim lim .x c x c

f x f x→ →− +

( ) ≠ ( )

For this function,

and so there is a jump discontinuity at x = 7.(D) This option is incorrect. The type of discontinuity was misidentified.

Since lim lim . ,x x

f x f x→ →− +

( ) = ( ) ≈10 10

0 5 there is no jump discontinuity at

x = 10. The function is defined at x = 10 and has a removable discontinuity there.

lim

limx

x

f x

f x→

→

−

+

( ) =

( ) =7

7

4

1


Question 84






(A) This option is incorrect. A misunderstanding of the Fundamental

Theorem of Calculus resulted in computing

(B) This option is incorrect. Computing f f1 5( ) + ′( ) shows a misunderstanding of the Fundamental Theorem of Calculus.

(C) This option is incorrect. A common error is to neglect to use the initial

condition and just calculate f x dx5 631

5( ) = +∫ .

This may arise from

a misunderstanding of the Fundamental Theorem of Calculus or just forgetting to add the initial condition after computing the definite integral with the calculator.

(D) This option is correct. By the Fundamental Theorem of Calculus,

f f f x dx5 11

5( ) − ( ) = ′( )∫ .

Therefore

f x dx F b F aa

b( ) = ( ) − ( )∫ .

f f x dx11

5( ) + ′′( )∫ .

f f x dx5 1 631

5( ) = ( ) + +∫

= + ≈π 24 672 27 814. . .


Question 85


2.3D Solve problems involving rates of change in applied contexts.

2.3D1: The derivative can be used to express information about rates of change in applied contexts.



(A) This option is incorrect. The focus was only on the rate at which people enter the building. This inequality says that the total number of people who have entered the building is increasing.

(B) This option is incorrect. Here the focus is only on the rate at which people enter the building. This inequality says that the rate of change of the number of people entering the building is increasing.

(C) This option is incorrect. The expression R t f t g t( ) = ( ) − ( ) indicates the rate of change of the number of people in the building (rate in minus rate out). The sign of R t( ) determines whether the number of people in the building is increasing or decreasing at time

t. This

inequality says that the number of people in the building is increasing.

(D) This option is correct. The expression R t f t g t( ) = ( ) − ( ) indicates the rate of change of the number of people in the building (rate in minus rate out). If the derivative ′( ) = ′( ) − ′( )R t f t g t is positive, then R t( ) is increasing, i.e. the rate of change of the number of people in the building is increasing.


Question 86



2.3C1: The derivative can be used to solve rectilinear motion problems involving position, speed, velocity, and acceleration.



(A) This option is incorrect. The meaning of the sign of the velocity may have been misinterpreted.

(B) This option is correct. At t = 1, the graph of v is positive. In addition, the graph of v is increasing at t = 1, which means that ′( )v 1 is positive. Alternatively, v 1 1 912( ) = . and ′( ) =v 1 1 617. .

Since velocity is the derivative of position, a positive value of v at t = 1 means that x is increasing, i.e. the particle is moving to the right. Since acceleration is the derivative of velocity, a positive value of ′v at t = 1 means that the particle is moving with positive acceleration.

(C) This option is incorrect. Both claims are reversed.(D) This option is incorrect. The meaning of the sign of the derivative of

velocity may have been misinterpreted.

1t

v


Question 87


2.3A Interpret the meaning of a derivative within a problem.

2.3A1: The unit for f x′( ) is the unit for

f divided by

the unit for x.



(A) This option is incorrect. This is the units of the dependent variable only.(B) This option is incorrect. This is the units of the independent variable

only.(C) This option is incorrect. The dependent and independent variables were

inverted. While in many problems time is often the independent variable, in this model time is specified as a function of the air pressure and hence is the dependent variable. That is why it is important to understand the functional notation t f p= ( ).

(D) This option is correct. The derivative of the function t f p= ( ) is the

limit of the difference quotient ΔΔ

tp . The units on the dependent

variable, t, in the numerator is hours and the units on the independent variable, p, in the denominator is psi. The rate of change, therefore, has units in hours per psi.


Question 88






(A) This option is incorrect. A common error is to answer the question as if the graph drawn on the calculator were the graph of the function f, not the graph of the derivative ′f . The function ′f is increasing between its local minimum and its local maximum, and this interval appears to match the location of those two points. The precise values can be found by solving for where ′′( ) =f x 0.

(B) This option is incorrect. There may have been confusion about which property of the derivative provides information about where the function is increasing. If one believes that the desired property is where the graph of ′f is concave up, then these two intervals appear to work. The precise values of the endpoints can be found by solving for where ′′′( ) =f x 0.

(C) This option is correct. The function f

will be increasing where ′f is positive. Graph ′f on an interval that includes at least − ≤ ≤2 1x since that covers the choices in options (A) and (B).

The graph of ′f is always positive on the interval in the window. Thegraph therefore rules out options (A), (B), and (D). The best choice is therefore (C).

[One cannot verify graphically that ′f is positive for − < <∞ ∞x . The denominator in the formula for ′f is always positive. The numerator is clearly positive for x ≥ 0 . For x < 0 , observe that the graph of 2e x−

is concave up and therefore lies above the tangent line at x = 0. This tangent line is y x= −2 2 . Hence 2 2 2e x xx− > − > − since x < <0 2. ]

(D) This option is incorrect. There may have been confusion about which sign of the derivative provides information about the increasing behavior of the function. Or the observation that there are no zeros of ′f might incorrectly suggest that there are no intervals to investigate.

1

x

y

−1−2−3 1 2 3


Question 89


1.2B Determine the applicability of important calculus theorems using continuity.

1.2B1: Continuity is an essential condition for theorems such as the Intermediate Value Theorem, the Extreme Value Theorem, and the Mean Value Theorem.



(A) This option is incorrect. The number of times that f x( ) = 4 may have been confused with the number of times that x = 4, or it may have been thought that the Intermediate Value Theorem only applies to the interval a b,[ ] when f a f b( ) ≥ ≥ ( )4 , namely on 4 6, .[ ]

(B) This option is incorrect. It may have been thought that the Intermediate Value Theorem only applies to the interval a b,[ ] when f a f b( ) ≤ ≤ ( )4 , namely on 0 4,[ ] and 8 13, .[ ]

(C) This option is correct. Because f is continuous, the Intermediate Value Theorem will imply that f x( ) will equal 4 on an interval a b,[ ] where 4 is between f a( ) and f b( ). According to the table, this happens for the intervals 0 4, ,[ ] 4 6, ,[ ] and 8 13, .[ ] So there are at least three values of x for which the Intermediate Value Theorem applies and f x( ) = 4.

(D) This option is incorrect. It may have been thought that an additional value of x occurs on 6 13,[ ] because f f6 4 13( ) < < ( ). But this x-value may be the same as the one guaranteed by the Intermediate Value Theorem on the interval 8 13, .[ ]


Question 90


2.2B Recognize the connection between differentiability and continuity.

2.2B2: If a function is differentiable at a point, then it is continuous at that point.



(A) This option is incorrect. It does not recognize that statement II is true.(B) This option is correct. By the Fundamental Theorem of Calculus,

h h h x dx3 11

3( ) = −( ) + ′( )

−∫ . Since the region between the graph of ′h

and the x-axis has more area above the axis than below the axis,

the definite integral ′( )−∫ h x dx

1

3 is positive. Therefore h h3 1( ) > −( )

and statement I is false. The graph shows that derivative of h exists on the interval −( )1 3, . That means that the function h is differentiable on −( )1 3, and is therefore continuous on that interval. In particular, h is continuous at x = 1, so statement II must be true. Because h is continuous at x = 1 and ′( ) ≈h 1 0 5. , the graph of h has a tangent line at x = 1 that is not vertical. Locally near x = 1 the graph of h will look like that tangent line and thus there cannot be a vertical asymptote at x = 1. Therefore statement III is false.

(C) This option is incorrect. Statement I might be chosen by interpreting the given graph as the graph of h rather than its derivative. It might also be

chosen by recognizing that h h h x dx3 11

3( ) − −( ) = ′( )

−∫ but believing that

the symmetry of the areas within the semicircles results in a net change of 0 for h over the interval − ≤ ≤1 3x .

(D) This option is incorrect. Statement I might be chosen for the same reasons as in option (C). Statement III might be chosen by interpreting the given graph as the graph of h rather than its derivative and confusing vertical asymptote with vertical tangent, or by thinking that the vertical tangent on ′h at x = 1 results in a vertical asymptote on h at x = 1.


Answers to Multiple-Choice Questions Part A Part B

1 – B 2 – D 3 – B 4 – B 5 – C 6 – A 7 – B 8 – D 9 – A10 – D11 – C12 – D13 – A14 – D15 – B16 – D17 – A18 – D19 – B20 – C21 – B22 – C23 – C24 – C25 – B26 – A27 – A28 – C29 – C30 – D

76 – A77 – B78 – D79 – B80 – C81 – A82 – A83 – C84 – D85 – D86 – B87 – D88 – C89 – C90 – B


Section II: Free Response Free-response questions provide students with an opportunity to demonstrate their knowledge of correct mathematical reasoning and thinking. In most cases, an answer without supporting work will receive no credit; students are required to articulate the reasoning and methods that support their answer. Some questions will ask students to justify an answer or discuss whether a theorem can be applied. There are two free-response questions in Part A and four questions in Part B. Each part of the free-response section is timed, and students may use a graphing calculator only for Part A. During the timed portion for Part B of the free-response section, students are allowed to return to working on Part A questions, though without the use of a graphing calculator.

Curriculum Framework Alignment for Free-Response Question 1Learning Objective Essential Knowledge Mathematical

Practices for AP Calculus

2.1B Estimate derivatives.

2.1B1: The derivative at a point can be estimated from information given in tables or graphs.






MPAC 6: Communicating

2.3A Interpret the meaning of a derivative within a problem.

2.3A1: The unit for f x′( ) is the unit for f divided by the unit for x.

2.3A2: The derivative of a function can be interpreted as the instantaneous rate of change with respect to its independent variable.

2.3D Solve problems involving rates of change in applied contexts.

2.3D1: The derivative can be used to express information about rates of change in applied contexts.

3.2B Approximate a definite integral.

3.2B2: Definite integrals can be approximated using a left Riemann sum, a right Riemann sum, a midpoint Riemann sum, or a trapezoidal sum; approximations can be computed using either uniform or nonuniform partitions.

3.4B Apply definite integrals to problems involving the average value of a function.

3.4B1: The average value of a function f over an interval

a b,[ ] is 1b a f x dx

a

b

−( )∫ .


Question 1

(a) ( ) ( ) ( )7 5 18.56 15.5 1.57 5 2− −≈ =−

′ =h hh in/min 1 : answer with units

(b) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) (

( ) ( ) ( ) ( )

10

02 0 2 5 2 5 7 5 7 10 7 10

2 10.0 3 15.5 2 18.5 3 20.0 163.5

≈ − + − + − + −

= + + + =∫ h t dt h h h h� � � )�

Because h is an increasing function, the right Riemann sum approximation is

greater than ( )10

0.∫ h t dt

1 : right Riemann sum 1 : approximation

3 : 1 : overestimate

with reason

⎧⎪⎪⎨⎪⎪⎩

(c) ( )10

0Average depth in tank 1 16.62410= =∫B g t dt in

( ) ( )10

0Average dep 1 1 163.5 16.35 in 16.624 in10th in tank 10< == ∫ h t dtA <

Therefore, the average depth of the water in tank B is greater than the average depth of the water in tank A.

1 : integral 1 : average depth

3 : in tank

1 : answer with reason

⎧⎪⎪⎨⎪⎪⎩

B

(d) ( ) 0 76 .88=′g

The depth of the water in tank B is increasing at time 6=t because ( ) 0.6′ >g

( ) 1 : uses 62 :

1 : answer with reason′⎧

⎨⎩

g

Scoring Guidelines for Free-Response Question 1




1.2B Determine the applicability of important calculus theorems using continuity.

1.2B1: Continuity is an essential condition for theorems such as the Intermediate Value Theorem, the Extreme Value Theorem, and the Mean Value Theorem.









2.3C1: The derivative can be used to solve rectilinear motion problems involving position, speed, velocity, and acceleration.

2.4A Apply the Mean Value Theorem to describe the behavior of a function over an interval.

2.4A1: If a function f is continuous over the interval a b,[ ] and differentiable

over the interval a b, ,( ) the Mean Value Theorem guarantees a point within that open interval where the instantaneous rate of change equals the average rate of change over the interval.

3.4C Apply definite integrals to problems involving motion.

3.4C1: For a particle in rectilinear motion over an interval of time, the definite integral of velocity represents the particle’s displacement over the interval of time, and the definite integral of speed represents the particle’s total distance traveled over the interval of time.


Question 2

(a) ( ) 0>Qa t for 0 3.963.< <t

The velocity of particle Q is increasing on the interval ( ]0, 3.963 .

{ 1 : interval2 :

1 : reason

(b) ( ) ( ) ( )

(

3

03 0

2 3.4901021.490

= +

= + −= −

∫Q Q Qxx t dtv

) ( )1 : integral

3 : 1 : uses 1 : answe

0r

⎧⎪⎨⎪⎩

Qx

(c) ( )Rx t is differentiable. ⇒ ( )Rx t is continuous on 1 3.≤ ≤t

( ) ( )3 1 8 4 23 1 2− −= =−

R Rx x

Therefore, by the Mean Value Theorem, there is at least one time t, 1 3,< <t for which ( ) ( ) 2.′ = =R Rx t v t

( ) ( ) 1 :2 : 1 : conclusion, using

3 1

M1

VT3−⎧

⎪−

⎪⎨⎩

R Rx x

(d) ( )3 1.490= −Qx and ( )3 1.682 (or 1.681)=Qv

Particle Q is moving to the right from 1.490.= −x

( )3 8=Rx and ( )3 2= −Rv Particle R is moving to the left from 8.=x

Therefore, the particles are moving toward each other.

{ 1 : answer2 :

1 : explanation






1.1C3: Limits of the

indeterminate forms 00

and

∞∞ may be evaluated using

L’Hospital’s Rule.







2.1C Calculate derivatives. 2.1C3: Sums, differences, products, and quotients of functions can be differentiated using derivative rules.



2.2A3: Key features of the graphs of f, f ′, and f ′′ are related to one another.




3.3A2: If f is a continuous function on the interval a b, ,[ ] then

ddx f t dt f x

a

x( )( ) = ( )∫ ,

where x is between a and b.

3.3A3: Graphical, numerical, analytical, and verbal representations of a function f provide information about the function g defined as

g x f t dta

x( ) = ( )∫ .


Question 3

(a) ( ) ( )′ =f x g x

The function f has critical points at 6= −x and 2.=x

f has a local minimum at 6= −x because ′f changes fromnegative to positive at that value.

f has a local maximum at 2=x because ′f changes from positive to negative at that value.

( ) ( ) 1 : 3 : 1 : critical points

1 : classifications with justification

′ =⎧⎪⎨⎪⎩

f x g x

(b) ( ) ( )0

80 4π

−= = +∫f g t dt 8 1 : answer

(c) ( ) ( ) ( )4 8

4lim 4 0

−

−

→−= − = =∫x

g t dtf x f

( )24

lim 4 0→−

+ =x

x x

Using L’Hospital’s Rule, ( ) ( ) ( )

24 4 4

2 1lim lim lim2 4 2 4 4 24→− →− →−

′= = = = −

+ + −+x x x

f x f x g xx xx x

{ 1 : L Hospital s Rule2 :

1 : answer’ ’

(d) ( )( ) ( ) ( )

( )

2

22

1

1

′+=

+′

−x g x g x x

xh x

2

( )( ) ( ) ( )( )( )

( )( ) ( )

2

22

1 1 1 1 2 1

11

1

2 3 3 2 34

′+ −=

+

− −= = −

′g

hg

( ){ 2 : 3 :

1 : answer′h x






1.1C1: Limits of sums, differences, products, quotients, and composite functions can be found using the basic theorems of limits and algebraic rules.






2.1C Calculate derivatives.

2.1C3: Sums, differences, products, and quotients of functions can be differentiated using derivative rules.

2.1C5: The chain rule is the basis for implicit differentiation.


2.1D2: Higher order derivatives are represented with a variety of notations. For y f x= ( ), notations for the second derivative include d ydx

2

2 , f x′′( ), and y′′.

Higher order derivatives can

be denoted d ydx

n

n or f xn( ) ( ).




3.5A2: Some differential equations can be solved by separation of variables.


Question 4

(a) ( )( 22= − +dy y xdx )1

( )2 12 = +−

⌠⎮⌡ ∫dy x dxy

3ln 2 3− = + +xy x C

30ln 3 0 ln 33= + + ⇒ =CC

Because ( )0 5,=y 2,>y so 2 2.− = −y y 3

32 3+

− =x x

y e3

32 3+

= +x x

y e

Note: this solution is valid for all real numbers.

1 : separation of variables 2 : antiderivatives

5 : 1 : constant of integration and uses initial condition

1 : solves for

⎧⎪⎪⎨⎪⎪⎩ y

Note: max 3 5 [1-2-0-0] if no constant of integration

Note: 0 5 if no separation of variables

(b) 3

33lim 22+

→−∞

⎛⎜ +⎜⎝

=⎠

x x

xe

⎞⎟⎟

1 : answer

(c) ( )

( )( )1

2

, 33 2 1 1 2+−=dy

dx =

( ) ( )( )2

22 1 2 2= + + −d y dy x y xdxdx

( )( )( ) ( )( )

,

22

1 32 2 1 1 3 2 2 6+ += − =d y

dx

Because ( )

2

21, 3

0>d ydx

and 2

2d ydx

is continuous, the graph of

( )=y f x is concave up at the point ( )1, 3 .

( )

2

21, 3

2 : 3 :

1 : concave up with reason

⎧⎪⎨⎪⎩

yddx





1.1A(a) Express limits symbolically using correct notation.

1.1A2: The concept of a limit can be extended to include one-sided limits, limits at infinity, and infinite limits.







1.2A1: A function f is continuous at x c= provided that f c( ) exists, lim

x cf x

→( )

exists, and lim .

x cf x f c

→( ) = ( )

2.1C Calculate derivatives.

2.1C2: Specific rules can be used to calculate derivatives for classes of functions, including polynomial, rational, power, exponential, logarithmic, trigonometric, and inverse trigonometric.

2.2B Recognize the connection between differentiability and continuity.

2.2B2: If a function is differentiable at a point, then it is continuous at that point.


3.2C2: Properties of definite integrals include the integral of a constant times a function, the integral of the sum of two functions, reversal of limits of integration, and the integral of a function over adjacent intervals.

3.2C3: The definition of the definite integral may be extended to functions with removable or jump discontinuities.



f x dx F b F aa

b( ) = ( ) − ( )∫ .


Question 5

(a) ( ) )2

0 0lim lim 3 2 0

− −→ →= + =

x xf x x x (

( ) ( )2

0 0lim lim 2 3

+ +→ →= + =x

x xf x e

Therefore, ( )0

lim→x

f x does not exist, and f is not continuous at

0.=x

{ 1 : considers one-sided limits2 :

1 : answer with justification

(b) ( ) ( ) ( )22 6 2 6 2 2 10=−′ − = + = − + = −xf x

( ) ( )( )2 323

3 2 2 2=

′ = ==xx

e ef e6

( )( )

2 1 : 2 :

1 3:′ −′

⎧⎨⎩

ff

(c)

( )0′f does not exist because f is not continuous at 0.=x 1 : explanation

(d) ( ) ( ) ( ) ( )

( ) ( )[ ] ( )

1 0 12 21 1 0

1203 21

02

2

3 2 21

22

10 0 1 1 2 02 2

32 2

− −

−

= + + +

⎡ ⎤⎡ ⎤= + +⎢ ⎥⎣ ⎦ ⎣ ⎦

⎡⎛ ⎞ ⎤+ − − + + + − +⎜ ⎟⎢⎣⎝ ⎠ ⎦

= +

=

+

=

∫ ∫ ∫ t

t

f t dt t t dtg

t

e dt

et t

e

e

⎥

1 : integrals4 : 2 : antiderivatives

1 : answer

⎧⎪⎨⎪⎩






2.3C3: The derivative can be used to solve optimization problems, that is, finding a maximum or minimum value of a function over a given.






3.1A Recognize antiderivatives of basic functions.

3.1A2: Differentiation rules provide the foundation for finding antiderivatives.


3.3A1: The definite integral can be used to define new functions; for example,

f x e dttx( ) = −∫

2

0.

3.3A2: If f is a continuous function on the interval a b, ,[ ]

then ddx f t dt f xa

x( )( ) = ( )∫ ,

where x is between a and b.



3.4A Interpret the meaning of a definite integral within a problem.

3.4A1: A function defined as an integral represents an accumulation of a rate of change.

3.4E Use the definite integral to solve problems in various contexts.

3.4E1: The definite integral can be used to express information about accumulation and net change in many applied contexts.

f x dx F b F aa

b( ) = ( ) − ( )∫ .


Question 6

(a) ( ) ( )2 2 22

00 02 15

3

1

0 26

5

4

⎡= − + = − +

− + =⎣

=⎦∫ ∫L t dt t dt t t⎤

26 hundred bees leave the hive during the time interval 0 2.≤ ≤t

1 : integral3 : 1 : antiderivative

1 : answer

⎧⎪⎨⎪⎩

(b)

The total number of bees, in hundreds, in the hive at time t is

( ) ( )(0

35 .−+ ∫ E x L x dx )t

( ) ( )( ) ( ) (((

( )

4 4 20 0

4 20

43 20

16 3 2 15

35 3 18 15

35 9 1

35 35

35 64 14

5

4 6055

− = − − − +

= + − + −

⎡ ⎤= + − + −⎣

+ +

= + − + −⎦

=

∫ ∫∫

E x L x dx x x x dx

x x dx

x x x

)))

55 hundred bees are in the hive at time 4.=t

1 : expression for total3 : 1 : antiderivative

1 : answer

⎧⎪⎨⎪⎩

(c) Let ( )B t be the total number of bees, in hundreds, in the hive at time t, for 0 4.≤ ≤t

( ) ( ) ( )( )0

35 −= + ∫t

E x L x xtB d

( ) ( ) ( ) ( )( )23 18 15 3 1 5′ = − = − + − = − − −B t E t L t t t t t

( ) 0 1, 5′ = ⇒ = =B t t t

t ( )B t 0 35 1 28 4 55

The minimum numbers of bees in the hive for 0 4≤ ≤t is 28 hundred bees.

( ) ( ) 1 : sets 3 : 1 : answer

1 : justificatio

0

n

−⎧⎪⎨⎪⎩

=E t L t



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